tff. — . — __ «, 

BOWSER'S MATHEMATICS. 

ACADEMIC ALGEBRA. With numerous Examples. 

COLLEGE ALGEBRA. With numerous Examples. 

PLANE AND SOLID GEOMETRY. With numerous Exer- 
cises. 

ELEMENTS OF PLANE AND SPHERICAL TRIGONOME- 
TRY. With numerous Examples. 

A TREATISE ON PLANE AND SPHERICAL TRIGONOME- 
TRY, and its applications to Astronomy and Geodesy. 

With numerous Examples. 

AN ELEMENTARY TREATISE ON ANALYTIC GEOMETRY, 
embracing Plane Geometry, and an Introduction to 
Geometry of Three Dimensions. 

AN ELEMENTARY TREATISE ON THE DIFFERENTIAL 
AND INTEGRAL CALCULUS. With numerous Exam- 
ples. 

AN ELEMENTARY TREATISE ON ANALYTIC MECHANICS. 

With numerous Examples. 

AN ELEMENTARY TREATISE ON HYDROMECHANICS. 

With numerous Examples. 

* 



A TREATISE 

ON 

PLANE AND SPHERICAL 

TKIGONOMETKY. 



AND ITS APPLICATIONS TO 



A Q 



TKOITOMY A^D GEODESY, 



NUMEROUS EXAMPLES. 



BY 

EDWARD A. BOWSER, LL.D., 

PBOEE6BOB OF MATHEMATICS AND ENGINEERING IN RUTGERS COLLEGE. 



BOSTOX, U.S.A. : 

PUBLISHED BY D. C. HEATH & CO. 

1902. 



/n 



2, 



Copyright, 1892, 
By E. A. BOWSER. 



By exchange 

Army & Navy Club 
JUN 2 2 1940 



PREFACE. 

.- 

The present treatise on Plane and Spherical Trigo- 
nometry is designed as a text-book for Colleges, Scien- 
tific Schools, and Institutes of Technology. The aim 
has been to present the subject in as concise a form as 
is consistent with clearness, to make it attractive and 
easily intelligible to the student, and at the same 
time to present the fullest course of Trigonometry 
which is usually given in the best Technological 
Schools. 

Considerable care has been taken to instruct the 
student in the theory and use of Logarithms, and 
their practical application to the solution of triangles. 
It is hoped that the work may commend itself, not 
only to those who wish to confine themselves to the 
numerical calculations which occur* <in. ; Trigonometry, 
but also to those who intend to -pursue the study of 
the higher mathematics. 

The examples are very numerous and are carefully 
selected. Many are placed in immediate connection 
with the subject-matter which they illustrate. The 
numerical solution of triangles has received much 
attention, each case being treated in detail. The 



IV PREFACE. 

examples at the ends of the chapters have been care- 
fully graded, beginning with those which are easy, 
and extending to those which are more and more diffi- 
cult. These examples illustrate every part of the sub- 
ject, and are intended to test, not only the student's 
knowledge of the usual methods of computation, but 
his ability to grasp them in the many forms they may 
assume in practical applications. Among these exam- 
ples are some of the most elegant theorems in Plane 
and Spherical Trigonometry. 

The Chapters on De Moivre's Theorem, and Astron- 
omy, Geodesy, and Potyedrons, will serve to introduce 
the student to some of the higher applications of 
Trigonometry, rarely found in American text-books. 

In writing this book, the best English and French 
authors have been consulted. I am indebted especially 
to the works of Todhunter, Casey, Lock, Hobson, 
Clarke, Eustis, Snowball, M'Clelland and Preston, 
Smith, and Serret. 

It remains for me to express my thanks to my col- 
leagues, Prof. R. W. Prentiss for reading the MS., 
and Mr. I. S. Upson for reading the proof-sheets. 

Any corrections or suggestions, either in the text 
or the examples, will be thankfully received. 

E. A. B. 

Rutgers College, 

New Brunswick, N. J., April, 1892. 



TABLE OF CONTENTS. 

*o*Ko« 

PART L 
PLANE TRIGONOMETRY. 



CHAPTER I. 

Measurement of Angles. 

art. PAGE 

1. Trigonometry 1 

2. The Measure of a Quantity 1 

3. Angles 2 

4. Positive and Negative Angles 3 

5. The Measure of Angles , 3 

6. The Sexagesimal Method 5 

7. The Centesimal or Decimal Method 6 

8. The Circular Measure 6 

9. Comparison of the Sexagesimal and Centesimal Measures . . 8 

10. Comparison of the Sexagesimal and Circular Measures 9 

11. General Measure of an Angle 11 

12. Complement and Supplement of an Angle 12 

Examples 13 



CHAPTER II. 
The Trigonometric Functions. 

13. Definitions of the Trigonometric Functions 16 

14. The Functions are always the Same for the Same Angle 18 

15. Functions of Complemental Angles 20 

16. Representation of the Functions by Straight Lines 20 

17. Positive and Negative Lines 23 

v 



Vi CONTENTS. 

ART. PAGE 

18. Functions of Angles of Any Magnitude 23 

19. Changes in Sine as the Angle increases from 0° to 360° 25 

20. Changes in Cosine as the Angle increases from 0° to 360°. . . 26 

21. Changes in Tangent as the Angle increases from 0° to 360°. 27 

22. Table giving Changes of Functions in Four Quadrants 28 

. 23. Relations between the Functions of the Same Angle 29 

24. Use of the Preceding Formulae 30 

25. Graphic Method of finding the Functions in Terms of One. . 30 

26. To find the Trigonometric Functions of 45° 31 

27. To find the Trigonometric Functions of 60° and 30° 31 

28. Reduction of Functions to 1st Quadrant 33 

29. Functions of Complemental Angles 34 

30. Functions of Supplemental Angles 34 

31. To prove sin (90° + A) = cos A, etc 35 

32. To prove sin (180° + A) = - sin A, etc 35 

33. To prove sin ( — A) = — sin A, etc 36 

34. To prove sin (270° + A) = sin (270° - A) = - cos A, etc 36 

35. Table giving the Reduced Functions of Any Angle 37 

36. Periodicity of the Trigonometric Functions 38 

37. Angles corresponding to Given Functions , . 39 

38. General Expression for All Angles with a Given Sine 40 

39. An Expression for All Angles with a Given Cosine 41 

40. An Expression for All Angles with a Given Tangent 41 

41. Trigonometric Identities 43 

Examples 44 



CHAPTER III. 
Trigonometric Functions of Two Angles. 

42. Fundamental Formulae 50 

43. To find the Values of sin ( x + y) and cos (x -f y) 50 

44. To find the Values of sin (x — y) and cos (x — y) 52 

45. Formulae for transforming Sums into Products 55 

46. Useful Formulae 56 

47. Tangent of Sum and Difference of Two Angles 57 

48. Formulae for the Sum of Three or More Angles 58 

49. Functions of Double Angles 60 

50. Functions of 3 x in Terms of the Functions of x 61 

51. Functions of Half an Angle 63 

52. Double Values of Sine and Cosine of Half an Angle 63 



CONTENTS. vii 

ART. PAGE 

53. Quadruple Values of Sine and Cosine of Half an Angle 65 

54. Double Value of Tangent of Half an Angle . 6Q 

55. Triple Value of Sine of One-third an Angle 67 

56. Find the Values of the Functions of 22J° 69 

57. Find the Sine and Cosine of 18° 69 

58. Find the Sine and Cosine of 36° 70 

59. If A + B + C = 180°, to find sin A + sin B + sin C, etc 70 

60. Inverse Trigonometric Functions 72 

61. Table of Useful Formulae 75 

Examples 77 



CHAPTER IV. 
Logarithms and Logarithmic Tables. — Trigonometric Tables. 

62. Nature and Use of Logarithms 87 

63. Properties of Logarithms 87 

64. Common System of Logarithms 91 

65. Comparison of Two Systems of Logarithms 93 

66. Tables of Logarithms 95 

67. Use of Tables of Logarithms of Numbers 98 

68. To find the Logarithm of a Given Number 99 

69. To find the Number corresponding to a Given Logarithm . . . 102 
69a. Arithmetic Complement 103 

70. Use of Trigononletric Tables 105 

71. Use of Tables of Natural Trigonometric Functions 106 

72. To find the Sine of a Given Angle 106 

73. To find the Cosine of a Given Angle 106 

74. To find the Angle whose Sine is Given. 108 

75. To find the Angle whose Cosine is Given. 108 

76. Use of Tables of Logarithmic Trigonometric Functions 110 

77. To find the Logarithmic Sine of a Given Angle 112 

78. To find the Logarithmic Cosine of a Given Angle 112 

79. To find the Angle whose Logarithmic Sine is Given 114 

80. To find the Angle whose Logarithmic Cosine is Given 115 

81. Angles near the Limits of the Quadrant 116 

Examples 117 



Vlll CONTENTS. 



CHAPTER V. 

Solution of Trigonometric Equations, 
art. PAGE 

82. Trigonometric Equations 126 

83. To solve m sin = a, m cos = b 128 

84. To solve a sin 0+6 cos = c 129 

85. To solve sin (« -f x) = m sin x 131 

86. To solve tan (<t -f x) = m tan x 132 

87. To solve tan (« + a;) tan x = m 133 

88. To solve m sin (0 + x) = a, m sin (0 + x) = 6 134 

89. To solve x cos cc -\- y sin a = m, x sin a — y cos a = n 135 

90. Adaptation to Logarithmic Computation 135 

91. To solve r cos cos 6 = a, r cos sin 6 = b, r sin = c 137 

92. Trigonometric Elimination 138 

Examples 140 

CHAPTER VI. 

Relations between the Sides of a Triangle and the Functions 
of its Angles. 

93. Formulae 146 

94. Right Triangles 146 

95. Oblique Triangles — Law of Sines 147 

96. Law of Cosines 148 

97. Law of Tangents 149 

98. To prove c = a cos B + b cos A 149 

99. Functions of Half an Angle in Terms of the Sides 150 

100. To express the Sine of an Angle in Terms of the Sides 152 

101. Expressions for the Area of a Triangle 153 

102. Inscribed Circle 154 

103. Circumscribed Circle 154 

104. Escribed Circle 155 

105. Distance between the In-centre and the Circumcentre 155 

106. To find the Area of a Cyclic Quadrilateral 157 

Examples 159 



CHAPTER VII. 
Solution of Triangles. 

107. Definitions 165 

108. Four Cases of Right Triangles 165 



CONTENTS. ix 

ART. PAGE 

109. Case I. — Given a Side and the Hypotenuse 166 

110. Case II. — Given an Acute Angle and the Hypotenuse 1G7 

111. Case III. — Given a Side and an Acute Angle 168 

112. Case IV. — Given the Two Sides 169 

IIS. When a Side and the Hypotenuse are nearly Equal 169 

114. Four Cases of Oblique Triangles 172 

115. Case I. — Given a Side and Two Angles 172 

116. Case II. — Given Two Sides and the Angle opposite One of them, 173 

117. Case III. — Given Two Sides and the Included Angle 176 

118. Case IV. — Given the Three Sides 177 

119. Area of a Triangle 180 

120. Heights and Distances — Definitions 181 

121. Heights of an Accessible Object 182 

122. Height and Distance of an Inaccessible Object 182 

123. An Inaccessible Object above a Horizontal Plane 184 

124. Object observed from Two Points in Same Vertical Line 185 

125. Distance between Two Inaccessible Objects 186 

126. The Dip of the Horizon 186 

127. Problem of Pothenot or of Snellius 188 

Examples 189 

CHAPTER VIII. 
Construction or Logarithmic and Trigonometric Tables. 

128. Logarithmic and Trigonometric Tables 204 

129. Exponential Series 204 

130. Logarithmic Series 205 

131. Computation of Logarithms 207 

132. Sin0 and tan are in Ascending Order of Magnitude 208 

133. The Limit of — is Unity 209 

d 

134. Limiting Values of sin 6 and cos 6 209 

135. To calculate the Sine and Cosine of 10" and of 1' 211 

136. To construct a Table of Natural Sines and Cosines 213 

137. Another Method 213 

138. The Sines and Cosines from 30° to 60° 214 

139. Sines of Angles Greater than 45° 215 

140. Tables of Tangents and Secants 215 

141. Formulae of Verification 216 

142. Tables of Logarithmic Trigonometric Functions 217 

143. The Principle of Proportional Parts. 218 



X CONTENTS. 

ART. PAGE 

144. To prove the Rule for the Table of Common Logarithms . . . 218 

145. To prove the Rule for the Table of Natural Sines 219 

146. To prove the Rule for a Table of Natural Cosines 219 

147. To prove the Rule for a Table of Natural Tangents 220 

148. To prove the Rule for a Table of Logarithmic Sines 221 

149. To prove the Rule for a Table of Logarithmic Cosines 222 

150. To prove the Rule for a Table of Logarithmic Tangents 222 

151. Cases of Inapplicability of Rule of Proportional Parts 223 

152. Three Methods to replace the Rule of Proportional Parts . . . 224 
Examples 226 



CHAPTER IX. 
De Moivre's Theorem. — Applications. 

153. De Moivre's Theorem 229 

p 

154. To find all the Values of (cos + V- 1 sin 0) ? 231 

155. To develop cos n0 and sin nd in Powers of sin and cos 0. . . . 233 

156. To develop sin and cos in Series of Powers of 234 

157. Convergence of the Series 235 

158. Expansion of cos n in Terms of Cosines of Multiples of 0. . . 235 

159. Expansion of sin n in Terms of Cosines of Multiples of 0. . . 236 

160. Expansion of sin n in Terms of Sines of Multiples of 237 

161. Exponential Values of Sine and Cosine 238 

162. Gregory's Series 239 

163. Euler's Series 240 

164. Machin's Series 241 

165. Given sin0 = x sin (0 + a) ; expand in Powers of x 242 

166. Given tan x = n tan ; expand x in Powers of n 242 

167. Resolve x n — 1 into Eactors 243 

168. Resolve x n + 1 into Factors 244 

169. Resolve x 2n - 2 x n cos + 1 into Eactors 245 

170. De Moivre's Property of the Circle 247 

171. Cote's Properties of the Circle 248 

172. Resolve sin into Eactors , 248 

173. Resolve cos into Eactors 250 

174. Sum the Series sin a + sin( a-\- &) + etc 251 

175. Sum the Series cos a + cos(rc + £) + etc 252 

176. Sum the Series sin™ a + sin w (a + /3) + etc 252 

177. Sum the Series sin a — sin(tt + /3) + etc 254 

178. Sum the Series cosec 6 -f cosec 2 + cosec 40 + etc 254 



CONTENTS. XI 



ART. PAGE 

6 6 

179. Sum the Series tan + l tan - + 1 tan + etc 255 

2 2 * 4 

180. Sum the Series sin a + x sin ( a+ £) + etc 255 

181. Summation of Infinite Series 256 

Examples 257 



PART II 
SPHERICAL TRIGONOMETRY. 



CHAPTER X. 
Formula relative to Spherical Triangles. 

182. Spherical Trigonometry 267 

183. Geometric Principles , 267 

184. Fundamental Definitions and Properties 268 

185. Formulae for Eight Spherical Triangles 270 

186. Napier's Rules 272 

187. The Species of the Parts 273 

188. Ambiguous Solution 274 

189. Quadrantal Triangles . 274 

190. Law of Sines 276 

191. Law of Cosines 277 

192. Relation between a Side and the Three Angles 278 

193. To find the Value of cot a sin 6, etc 279 

194. Useful Formulas . . 280 

195. Formulae for the Half Angles 281 

196. Formulae for the Half Sides 284 

197. Napier's Analogies , 286 

198. Delambre's (or Gauss's) Analogies ... 287 

Examples , 288 

CHAPTER XL 

Solution of Spherical Triangles. 

199. Preliminary Observations 297 

200. Solution of Right Spherical Triangles 297 



xii CONTENTS. 

ART. PAGE 

201. Case I. — Given the Hypotenuse and an Angle 298 

202. Case II. — Given the Hypotenuse and a Side 299 

203. Case III. — Given a Side and the Adjacent Angle 300 

204. Case IV. — Given a Side and the Opposite Angle 301 

205. Case V. — Given the Two Sides 302 

206. Case VI. — Given the Two Angles 302 

207. Quadrantal and Isosceles Triangles „ 303 

208. Solution of Oblique Spherical Triangles 304 

209. Case I. — Given Two Sides and the Included Angle 305 

210. Case II. — Given Two Angles and the Included Side 307 

211. Case III. — Given Two Sides and One Opposite Angle 309 

212. Case IV. — Given Two Angles and One Opposite Side 312 

213. Case V. — Given the Three Sides 313 

214. Case VI. — Given the Three Angles 314 

Examples 316 



CHAPTER XII. 
The In-Circles and Ex-Circles. — Areas. 

215. The Inscribed Circle ■ 324 

216. The Escribed Circles 325 

217. The Circumscribed Circle 326 

218. Circumcircles of Colunar Triangles 328 

219. Areas of Triangles. — Given the Three Angles 329 

220. Areas of Triangles. — Given the Three Sides 330 

221. Areas of Triangles. — Given Two Sides and the Included Angle, 331 
Examples 332 

CHAPTER XIII. 
Applications of Spherical Trigonometry. 

222. Astronomical Definitions 338 

223. Spherical Coordinates 339 

224. Graphic Representation of the Spherical Coordinates 341 

225. Problems 342 

226. The Chorda] Triangle 346 

227. Legendre's Theorem 348 

228. Roy's Rule 350 

229. Reduction of an Angle to the Horizon 352 



COy TESTS. xiii 

ART. PAGE 

230. Small Variations in Parts of a Spherical Triangle 353 

231. Inclination of Adjacent Faces of Polyedrons 356 

232. Volume of Parallelopiped 357 

233. Diagonal of a Parallelopiped 358 

234. Table of Formulae in Spherical Trigonometry 359 

Examples 362 



TREATISE ON TRIGONOMETRY. 



PART L 
PLANE TRIGONOMETRY. 

CHAPTER I. 

MEASUKEMENT OP ANGLES. 

* 

1. Trigonometry is that branch, of mathematics which 
treats (1) of the solution of plane and spherical triangles, 
and (2) of the general relations of angles and certain func- 
tions of them called the trigonometric functions. 

Plane Trigonometry comprises the solution of plane trian- 
gles and investigations of plane angles and their functions. 

Trigonometry was originally the science which treated only of the 
sides and angles of plane and spherical triangles ; but it has been 
recently extended so as to include the analytic treatment of all theo- 
rems involving the consideration of angular magnitudes. 

2. The Measure of a Quantity. — All measurements of 
lines, angles, etc., are made in terms of some fixed standard 
or unit, and the measure of a quantity is the number of times 
the quantity contains the unit. 

Tt is evident that the same quantity will be represented 
by different numbers when different units are adopted. For 
example, the distance of a mile will be represented by the 
number 1 when a mile is the unit of length, by the number 
1760 when a yard is the unit of length, by the number 5280 
when a foot is the unit of length, and so on. In like man- 

1 



2 PLANE TRIGONOMETRY. 

ner, the number expressing the magnitude of an angle will 
depend on the unit of angle. 

EXAMPLES. 

1. What is the measure of 2-J- miles when a yard is the 
unit ? 

2£ miles = f X 1760 yards 

= 4400 yards = 4400 x 1 yard. 
.\ the measure is 4400 when a yard is the unit. 

2. What is the measure of a mile when a chain of 66 feet 
is the unit ? Ans. 80. 

3. What is the measure of 2 acres when a square whose 
side is 22 yards is the unit ? Ans. 20. 

4. The measure of a certain field is 44 and the unit is 
1100 square yards ; express the area of the field in acres. 

Ans. 10 acres. 

5. If 7 inches be taken as the unit of length, by whab 
number will 15 feet 2 inches be represented ? Ans. 26. 

6. If 192 square inches be represented by the number 12, 
what is the unit of linear measurement ? Ans. 4 inches. 

3. Angles. — An angle is the opening between two straight 
lines drawn from the same point. The point is called the 
vertex of the angle, and the straight lines are called the sides 
of the angle. 

An angle may be generated by revolving a line from coin- 
cidence with another line about a fixed point. The initial 
and final positions of the line are the , q 
sides of the angle; the amount of 
revolution measures the magnitude of 
the angle ; and the angle may be traced 
out by any number of revolutions of 
the line. O A 

Thus, to form the angle AOB, OB may be supposed to 
have revolved from OA to OB ; and it is obvious that OB 






THE CIRCULAR MEASURE. 3 

may go on revolving until it comes into the same position 
OB as many times as we please ; the angle AOB, having the 
same bounding lines OA and OB, may therefore be greater 
than 2, 4, 8, or any number of right angles. 

The line OA from which OB moves is called the initial 
line , and OB in its final position, the terminal line. The 
revolving line OB is called the generatrix. The point is 
called the origin, vertex, or pole. 

4. Positive and Negative Angles. — We supposed in Art. 
3 that OB revolved in the direction opposite to that of the 
hands of a watch. But angles may, of course, be described 
by a line revolving in the same direction as the hands of a 
watch, and it is often necessary to distinguish between the 
two directions in which angles may be measured from the 
same fixed line. This is conveniently effected by adopting 
the convention that angles measured in one direction shall 
be considered positive, and angles measured in the opposite 
direction, negative. In all branches of mathematics angles 
described by the revolution of a straight line in the direc- 
tion opposite to that in which the hands of a watch move 
are usually considered positive, and all angles described by 
the revolution of a straight line in the same direction as the 
hands of a watch move are considered negative. 

Thus, the revolving line OB starts from the initial line 
OA. When it revolves in the B 
direction contrary to that of the 
hands of a w^atch, and comes into 
the position OB, it traces out the / \ 

positive angle AOB (marked + a) ; 
and when it revolves in the same 
direction as the hands of a watch, 
it traces the negative angle AOB (marked — b). 

The revolving line is always considered positive. 

5. The Measure of Angles. — An angle is measured by 
the arc of a circle whose centre is at the vertex of the 





4 PLANE TRIGONOMETRY. 

angle and whose ends are on the sides of the angle (Geom., 
Art. 236). 

Let the line OP of fixed length generate an angle by 
revolving in the positive direction 
round a fixed point from an initial 
position OA. Since OP is of constant 
length, the point P will trace out the 
circumference ABA'B' whose centre 
is 0. The two perpendicular diam- 
eters A A' and BB' of this circle will 
inclose the four right angles AOB, 
BOA', A'OB', and B'OA. 

The circumference is divided at the points A, B, A', B f 
into four quadrants, of which 

AB is called the first quadrant. 
BA' " " " second quadrant. 
A'B' " " " third quadrant. 
g^ u u u fourth quadrant. 

In the? figure, the angle AOP 1? between the initial line 
OA and the revolving line 0P 1? is less than a right angle, 
and is said to be an angle in the first quadrant. AOP 2 is 
greater than one and less than two right angles, and is said to 
be an angle in the second quadrant. AOP 3 is greater than two 
and less than three right angles, and is said to be an angle 
in the third quadrant. AOP 4 is greater than three and less 
than four right angles, and is said to be an angle in the 
fourth quadrant. 

When the revolving line returns to the initial position 
OA, the angle AOA is an angle of four right angles. By- 
supposing OP to continue revolving, the angle described 
will become greater than an angle of four right angles. 
Thus, when OP coincides with the lines OB, OA', OB', OA, 
in the second revolution, the angles described, measured 
from the beginning of the first revolution, are angles of five 
right angles, six right angles, seven right angles, eight right 



THE SEXAGESIMAL METHOD. 5 

angles, respectively, and so on. By the continued revolu- 
tion of OP the angle between the initial line OA and the 
revolving line OP may become of any magnitude whatever. 

In the same way OP may revolve in the negative direc- 
tion about any number of times, generating a negative 
angle ; and this negative angle may obviously have any 
magnitude whatever. 

The angle AOP may be the geometric representative of 
any of the Trigonometric angles formed by any number of 
complete revolutions, either in the positive direction added 
to the positive angle AOP, or in the negative direction added 
to the negative angle AOP. In all cases the angle is said to 
be in the quadrant indicated by its terminal line. 

There are three methods of measuring angles, called 
respectively the Sexagesimal, the Centesimal, and the Cir- 
cular methods. 

6. The Sexagesimal Method. — This is the method in 
general use. In this method the right angle is divided into 
90 equal parts, each of which is called a degree. Each 
degree is subdivided into 60 equal parts, each of which is 
called a minute. Each minute is subdivided into 60 equal 
parts, each of which is called a second. Then the magni- 
tude of an angle is expressed by the number of degrees, 
minutes, and seconds which it contains-. Degrees, minutes, 
and seconds are denoted respectively by the symbols °, ', " : 
thus, to represent 18 degrees, 6 minutes, 34.58 seconds, we 
write 

18° 6' 34".58. 

A degree of arc is ^\-^ of the circumference to which the 
arc belongs. The degree of arc is subdivided in the same 
manner as the degree of angle. 

Then 1 circumference = 360° = 21600' = 1296000". 
1 quadrant or right angle = 90°. 

Instruments used for measuring angles are subdivided 
accordingly. 



6 PLANE TRIGONOMETRY. 

7. The Centesimal or Decimal Method. — In this method 
the right angle is divided into 100 equal parts, each of 
which is called a grade. Each grade is subdivided into 100 
equal parts, each of which is called a minute. Each minute 
is subdivided into 100 equal parts, each of which is called 
a second. The magnitude of an angle is then expressed by 
the number of grades, minutes, and seconds which it con- 
tains. Grades, minutes, and seconds are denoted respect- 
ively by the symbols g , \ VN : thus, to represent 34 grades, 
48 minutes, 86.47 seconds, we write 

34 g 4gv 86 N \47. 

The centesimal or decimal method was proposed by the French 
mathematicians in the beginning of the present century. But 
although it possesses many advantages over the established method, 
they were not considered sufficient to counterbalance the enormous 
labor which would have been necessary to rearrange all the mathe- 
matical tables, books of reference, and records of observations, which 
would have to be transferred into the decimal system before its 
advantages could be felt. Thus, the centesimal method has never been 
used even in France, and in all probability never will be used in prac- 
tical work. 



8. The Circular Measure. — The unit of 
circular measure is the angle subtended at 
the centre of a circle by an arc equal in 
length to the radius. 

This unit of circular measure is called a 
radian. 

Let O be the centre of a circle whose radius is r, 

Let the arc AB be equal to the radius 
OA = r. 

Then, since angles at the centre of a 
circle are in the same ratio as their 
intercepted arcs (Geom., Art. 234), and 
since the ratio of the circumference of 
a circle to its diameter is tt = 3.14159265 
(Geom., Art. 436), 






THE CIRCULAR MEASURE. 7 

.\ angle AOB : 4rt. angles : : arc AB : circumference, 

: : r : 2vr : : 1 : 2w. 

-, A ,-vd 4 rt. angles 2 rt. angles 
.-. angle AOB = — ^ — = § 

.-. a radian = angle AOB = *f?° = 57°.2957795 

= 3437'.74677 = 206264".806. 

Therefore, the radian is the same for all circles, and 
= 57°.2957795. 

Let ABP be any circle ; let the angle 
AOB be the radian; and let AOP be 
any other angle. 

Then arc AB = radius OA. 

.-. angle AOP : angle AOB 

: : arc AP : arc AB ; 
or angle AOP : radian : : arc AP : radius. 

.-. angle AOP = ; — x radian. 

radius 

The measure of any quantity is the number of times it 
contains the unit of measure (Art. 2). 

,\ the circular measure of angle AOP = : 

radius 

Note 1. — The student will notice that a radian is a little less than an angle of an 
equilateral triangle, i.e., of 60°. 

Angles expressed in circular measure are usually denoted by Greek letters, a, j3, 
y, ..., </>, 9, i//, .... 

The circular measure is employed in the various branches of Analytical Mathe- 
matics, in which the angle under consideration is almost always expressed by a 
letter. 

Note 2. — The student cannot too carefully notice that unless an angle is obvi- 
ously referred to, the letters a, /3, ..., 0, </>, ... stand for mere numbers. Thus, rr stands 
for a number, and a number only, viz., 3.14159 ..., but in the expression ' the angle 
7r,' that is, • the angle 3.14159 ...,' there must be some unit understood. The unit 
understood here is a radian, and therefore * the angle it ' stands for ' n radians' or 
3.14159 ... radians, that is, two right angles. 

Hence, when an angle is referred to, it is a very convenient abbreviation for 
two right angles. 

So also ' the angle a or 6* means ' a radians or 9 radians.' 

The units in the three systems, when expressed in terms 
of one common standard, two right angles, stand thus : 



8 PLANE TRIGONOMETRY, 

The unit in the Sexagesimal Method - ■ of 'J right angles. 



Centesimal 


u 


200 


u 


u 


Circular 


u 


= 1 .. .. 

7T 


a 


« 



it i>, i;, and 6 denote the number o( degrees, grades, and 
radians respectively in :w\\ angle, then 

P (; ° (I) 

ISO LW ? V 

because eaoh fraction is the ratio of the angle to two right 

angles. 

9. Comparison of the Sexagesimal and Centesimal Meas- 
ures of an Angle. Although the eentesunal method was 

never In general use among mathematicians, and is now 
totally abandoned everywhere, yet it still possesses some 

interest, as it shows the application of the deeiinal system 

to the measurement of angles. 

From (1) o( Art. S we have 

D G 

ISO LVO' 

.-. l>- ? (}, and G - ,0 P. 
10 9 

t. Express 49° IS 1 S6 H Ln eentesunal measure. 

First ezpr68S the angle in degrees and deeimals of a 

degree thus : 

80) 



60) 15' ,583 

49°. 25972 

10 

9) 492,5972 
54*738024 ••• , 
.-. 4»" US -i- ftf 80 x \24« 



COMPARISON OF MEASURES. 9 

2. Express 87* 2 s 25" in degrees, etc. 
First express the angle in grades and decimals of a grade 
thus: 

87g 2 V 25" =87*0225 
.9_ 

78.32025 
60 



19.215 
60 



12.9 

.-. 87*2 V 25"=78°19'12".9. 

Find the number of grades, minutes, and seconds in the 
following angles : 

3. 51° 4' 30". Ans. 56* 75 N X \ 

4. 45° 33' 3". 50* 61 x 20 v \37. 

5. 27° 15' 46". 30* 29 v 19 v \75 .... 

6. 157° 4' 9". 174*52 x 12 v \962-... 

Find the number of degrees, minutes, and seconds in the 
following angles : 

7. 19* 45 v 95 v \ Ans. 17° 30' 48".78. 

8. 124* 5 V 8 V \ 111° 38' 44".592. 

9. 55* 18 v 35 v \ 49° 39' 54".54. 

10. Comparison of the Sexagesimal and Circular Meas- 
ures of an Angle. 

From (1) of Art. 8 we have 

_d = e 

180 tt' 

t. 1800 , a Dtt 
,.D = — ,and0 = -- 



10 PLANE TRIGONOMETRY. 

EXAMPLES. 

1. Find the number of degrees in the angle whose circu- 



lar measure is ^. 




Here 


«4 


.-. D = 


180 v 1 90 
- x 9 — — 

7T Z 7T 




90x7 



= 28° 38' 10"if , 

where ^£ is used for ir. 

2. Find the circular measure of the angle 59° 52' 30 ". 
Express the angle in degrees and decimals of a degree 

thus : 

60 )52.5 
59.875 

.-. 6 = ^jl^7r = (.333. ..)tt =1.0453-... 

3. Express, in degrees, the angles whose circular measures 

7T 7T 7T 7T o 

are — , — , — , — , -q7t. 
2' 3 4 6 3 

Note 1. — The student should especially accustom himself to express readily in 
circular measure an angle which is given in degrees. 

4. Express in circular measure the following angles : 

60°, 22° 30', 11° 15', 270°. Arts. -, -, — , &. 

3 8' 16' 2 

5. Express in circular measure 3° 12 \ and find to seconds 
the angle whose circular measure is .8. 

Take it = — -\ Ans. — , 45° 49' 5"JU 

7 J 225 xl 

6. One angle of a triangle is 45°, and the circular measure 
of another is 1.5. Find the third angle in degrees. 

Ans. 49° 5' 27" T 3 T . 

Note 2. — Questions in which angles are expressed in different systems of meas- 
urement are easily solved by expressing each angle in right angles. 



GENERAL MEASURE OF AN ANGLE. 



11 



7. The sum of the measure of an angle in degrees and 
twice its measure in radians is 23|-; find its measure in 
degrees (?r = - 2 T 2 -). 

Let the angle contain x right angles. 

Then the measure of the an^le in decrees = 90 #. 



(( 


u u 


cc u 


radians 


IT 


X. 




■\ 90a 


+ ttX 


— 2?A • 








.\ 90a 


+ ^X 


163 

" 7 5 










.-. 652 a 


= 163, .• 


. X: 


_1 

4 


the 


angle is 


J of 90° 


= 22\\ 







8. The difference between two angles is -, and their sum 

° 9 

is 56° ; find the angles in degrees. Ans. 38°, 18°. 

11. General Measure of an Angle. — In Euclidian geom- 
etry and in practical applications of trigonometry, angles 
are generally considered to be less than two right angles ; 
but in the theoretical parts of mathematics, angles are 
treated as quantities which may be of any magnitude what- 
ever. 

Thus, when we are told that an angle is in some particu- 
lar quadrant, say the second (Art. 5), we know that the 
position in which the revolving line stops is in the second 
quadrant. But there is an unlimited number of angles 
having the same final position, OP. 

The revolving line OF may pass from 
OA to OP, not only by describing the 
arc ABP, but by moving through a whole 
revolution plus the arc ABP, or through 
any number of revolutions plus the arc 
ABP. 

For example, the final position of OP 
may represent geometrically all the fol- 
lowing angles : 




12 PLANE TRIGONOMETRY. 

Angle AOP = 130°, or 360° + 130°, or 720° + 130°, or 
- 360° + 130°, or - 720° + 130°, etc. 

Let A be an angle between and 90°, and let n be any 
whole number, positive or negative. Then 

(1) 2n x 180° + A represents algebraically an angle in the 

first quadrant. 

(2) 2n x 180° — A represents algebraically an angle in the 

fourth quadrant. 

(3) {2n + 1) 180° — A represents algebraically an angle in 

the second quadrant. 

(4) (2n + 1) 180° + A represents algebraically an angle in 

the third quadrant. 

In circular measure the corresponding expressions are 
(1) 2nir+6, (2) 2mr-0, (3) (2n+l)*— 6, (4) {2n + l)ir+6. 

EXAMPLES. 

State in which quadrant the revolving line will be after 
describing the following angles : 

(1) 120°, (2) 340°, (3) 490°, (4) - 100°, 
(o)-380° ; (6) fir, (7)10tt + : 



7T 



12. Complement and Supplement of an Angle or Arc. — 

The complement of an angle or arc is the remainder obtained 
by subtracting it from a right angle or 90°. 

The supplement of an angle or arc is the remainder 
obtained by subtracting it from two right angles or 180°. 
Thus, the complement of A is (90° — A). 

The complement of 190° is (90° - 190°) = -100°. 

The supplement of A is (180° - A). 

The supplement of 200° is (180° - 200°) = - 20°. 

The complement of f w is ( * — f -k j = — \ tt. 
The supplement of f tt is (tt — f tt) = \ tt. 



EXAMPLES. 13 

EXAMPLES. 

1. If 192 square inches be represented by the number 12, 
what is the unit of linear measurement ? Arts. 4 inches. 

2. If 1000 square inches be represented by the number 
40, what is the unit of linear measurement ? Ans. 5 inches. 

3. If 2000 cubic inches be represented by the number 16, 
what is the unit of linear measurement ? Ans. 5 inches. 

4. The length of an Atlantic cable is 2300 miles and the 
length of the cable from England to France is 21 miles. 
Express the length of the first in terms of the second as 
unit. Ans. 109|f 

5. Find the measure of a miles when b yards is the unit. 

A 1760a 

Ans. ■ — 

b 

6. The ratio of the area of one field to that of another is 
20 : 1, and the area of the first is half a square mile. Find 
the number of square yards in the second. Ans. 77440. 

7. A certain weight is 3.125 tons. What is its measure 
in terms of 4 cwt.? Ans. 15.625. 

Express the following 12 angles in centesimal measure : 



8. 


42° 15' 18". 


Ans. 46* 95\ 


9. 


63° 19' 17". 


70*35 v 70 v \98-... 


10. 


103° 15' 45". 


114* 7^ 6r\l. 


11. 


19° 0'18". 


21* ir 66 v \6. 


12. 


143° 9' 0". 


159* 5 V 55 V \5. 


13. 


300° 15' 58". 


333* 62 x 90 v \l234567890. 


14. 


27° 41' 51". 


308.775. 


15. 


67°.4325. 


74*925. 


16. 


8° 15' 27". 


9* IT 50 v \ 


17. 


97° 5' 15". 


107* 87 v 50". 


18. 


16° 14' 19". 


18* 4 X 29"-... 


19. 


132° 6'. 


146* 77 v 77 v \7. 



14 PLANE TRIGONOMETRY. 

Express the following 11 angles in degrees, minutes, and 
seconds : 

20. 105s 52 v 75 v \ Ans. 94° 58' 29".l. 

21. 82= 9 V 54 V \ 73° 53' 9".096. 

22. 70«15 v 92 v \ 63° 8' 35".808. 

23. 15& V 15 V \ 13° 30' 4".86. 

24. 154« 7 X 24 V \ 138° 39' 54".576. 

25. 324* 13 v 88 v \7. 291° 43' 29".93S8. 

26. 10« 42 v 50 x \ 9° 22' 57". 

27. 20« 77 v 50 x \ 18° 41' 51". 

28. 8 8 75\ 7° 52' 30". 

29. 170M5 V 35 V \ 153° 24' 29".34. 

30. 248 o^25 v \ 21° 36' 8".l. 

Express in circular measure the following angles : 

1453*- 



31. 315°, 24° 13'. Ans. \tt, 

143*- 



' 10800 



32. 95° 20', 12° 5' 4". 



270' 40500 



33. 221°, 1°, 57°.295. -, ,— , 1 radian. 

2 ' 8 180 

34. 120°, 45°, 270°. 2.09439, -, fir. 

35. 360°, 3-i- rt. angles. 2ir, \-k. 

Express in degrees, etc., the angles whose circular meas- 
ures are : 

36. fir, fir, \- Ans. 112°.5, 120°, — degrees. 

Q7 1 1 2 45 , 30 , 120 , 

**' ' 7> *' o' — degrees, — degrees, degrees. 

4 O 6 IT 7T 7T 

38. | .7854. 47° 43' 38" r \, 45°. 



EXAMPLES. 15 

39. 41, T L,r, 2.504 A,is. 257° 49' 43".39, 15°, 143°.468. 

40. .0234, 1.234, -• 1°20'27", 70°42'11", 38°11'50". 

o 

41. Find the number of radians in an angle at the centre 
of a circle of radius 25 feet, which intercepts an arc of 
S7|- feet. Ans. 1|. 

42. Find the number of degrees in an angle at the centre 
of a circle of radius 10 feet, which intercepts an arc of 
57rfeet. Ans. 90°. 

43. Find the number of right angles in an angle at the 
centre of a circle of radius 3^- inches, which intercepts an 
arc of 2 feet. Ans. 4f. 

44. Find the length of the arc subtending an angle of 
4J radians at the centre of a circle whose radius is 25 feet. 

Ans. 112i ft. 

45. Find the length of an arc of 80° on a circle of 4 feet 
radius. Ans. 5f|- ft. 

46. The angle subtended by the diameter of the Sun at 
the eye of an observer is 32' : find approximately the 
diameter of the Sun if its distance from the observer be 
90 000 000 miles. Ans. 838 000 miles. 

47. A railway train is travelling on a curve of half a mile 
radius at the rate of 20 miles an hour : through what angle 
has it turned in 10 seconds ? Ans. 6 T 4 T degrees. 

48. If the radius of a circle be 4000 miles, find the length 
of an arc which subtends an angle of 1" at the centre of the 
circle. Ans. About 34 yards. 

49. On a circle of 80 feet radius it was found that an 
angle of 22° 30' at the centre was subtended by an arc 
31 ft. 5 in. in length : hence calculate to four decimal places 
the numerical value of the ratio of the circumference of a 
circle to its diameter. Ans. 3.1416. 

50. Find the number of radians in 10" correct to four sig- 
nificant figures (use fff for «"> Ans - -00004848. 



16 PLANE TRIGONOMETRY. 




CHAPTER II. 
THE TKIGOHOMETEIC FUNCTIONS. 

13. Definitions of the Trigonometric Functions. — Let 

RAD be an angle ; in AD, one of the lines containing the 
angle, take any point B, and from B 
draw BC perpendicular to the other B^ 

line AR, thus forming a right triangle 
ABC, right-angled at C. Then denot- 
ing the angles by the capital letters A, 
B, C, respectively, and the three sides A b 

opposite these angles by the corresponding small italics, a, 
6, c } * we have the following definitions : 

a = opposite side ig called ^ ^ of the ftngle A 
c hypotenuse 

b ^ adjacent side ig ^^ the ^^ of the angle A 
c hypotenuse 

a ^ opposite side . g caUed the tangent of the angle A- 
b adjacent side 

6 ^ adjacent side . g called the cotangent of the angle A 
a opposite side 

£ a- _ZE — nn f is called the secant of the angle A. 
b adjacent side 

_ — — ZE__ i s called the cosecant of the angle A. 

a opposite side 

If the cosine of A be subtracted from unity, the remain- 
der is called the versed sine of A. If the sine of A be sub- 

* The letters a, b, c are numbers, being the number of times the lengths of the sides 
contain some chosen unit of length. 



TRIGONOMETRIC FUNCTIONS. 17 

tracted from unity, the remainder is called the coversed sine 
of A ; the latter term is hardly ever used in practice. 

The words sine, cosine, etc., are abbreviated, and the func- 
tions of an angle A are written thus: sin A, cos A, tan A, 
cot A, sec A, cosecA, vers A, covers A. 

The following is the verbal enunciation of these defini- 
tions : 

The sine of an angle is the ratio of the opposite side to the 

hypotenuse; orsinA = -. 

c 

The cosine of an angle is the ratio of the adjacent side to 

the hypotenuse; or cos A = — 

c 

The tangent of an angle is the ratio of the opposite side to 
the adjacent side; or tan A = — 

The cotangent of an angle is the ratio of the adjacent side to 

the opposite side; or cot A = -• 

a 

The secant of an angle is the ratio of the hypotenuse to the 

/> 
adjacent side; or sec A = — 

The cosecant of an angle is the ratio of the hypotenuse to the 

/> 

opposite side; or cosecA= — 

a 

The versed sine of an angle is unity minus the cosine of the 

angle; or vers A = 1 — cos A = 1 

c 

The coversed sine of an angle is unity minus the sine of the 

angle; or covers A = 1 — sin A = 1 

c 

These ratios are called Trigonometric Functions. The 
student should carefully commit them to memory, as upon 
them is founded the whole theory of Trigonometry. 

These functions are, it will be observed, not lengths, but 



18 PLANE TRIGONOMETRY. 

ratios of one length to another ; that is, they are abstract 
numbers, simply numerical quantities ; and they remain 
unchanged so long as the angle remains unchanged, as will 
be proved in Art. 14. 

It is clear from the above definitions that 

cosec A = , or sin A = — , 

sin A cosec A 

sec A = , or eosA = - 



cosA' sec A' 

tan A == , or cotA = 



cot A tan A 

The powers of the Trigonometric functions are expressed 
as follows : 

(sin A) 2 is written sin 2 A, 

(cos A) 3 is written cos 3 A, 
and so on. 

Note. — The student must notice that * sin A ' is a single symbol, the name of a 
number, or fraction belonging to the angle A. Also sin 2 A is an abbreviation for 
(sin A) 2 , i.e., for (sin A) x (sin A). Such abbreviations are used for convenience. 

14. The Trigonometric Functions are always the Same 
for the Same Angle. — Let BAD be 
any angle; in AD take P, P f , any 
two points, and draw PC, P'C per- fy 

pendicular to AB. Take P", any 
point in AB, and draw P"C" per- 
pendicular to AD. A c c ' p " 

Then the three triangles PAC, P'AC, P"AC" are equi- 
angular, since they are right-angled, and have a common 
angle at A : therefore they are similar. 

pc == fc; == f^ 

" AP~ AP'~ AP"' 

But each of these ratios is the sine of the angle A. Thus, 
sin A is the same whatever be the position of the point P on 
either of the lines containing the angle A. 




FUNCTIONS OF COMPLEMENTAL ANGLES. 19 

Therefore sin A is always the same. A similar proof 
may be given for each of the other functions. 
In the right triangle of xlrt. 13, show that 

a = c sin A = c cos B = b tan A = b cot B, 
b = a cot A = a tanB = c cos A = c sinB, 
c = a cosec A = a secB = b sec A = b cosecB. 

Note. — These results should be carefully noticed, as they ar^ of frequent use in 
the solution of right triangles and elsewhere. 

EXAMPLES. 

1. Calculate the value of the functions, sine, cosine, etc., 
of the angle A in the right triangles whose sides a, b, c are 
respectively (1) 8, 15, 17; (2) 40, 9, 41 ; (3) 196, 315, 371 ; 
(4) 480, 31, 481 ; (5) 1700, 945, 1945. 

Ans. (1) sin A = T 8 T , cos A = i|., tan A = y 8 ^, etc. ; 

(2) sin A = f|, cos A = -$ T , etc. ; 

(3) sin A = ||, tanA = ||, etc.; 

(4) sin A = fff, tan A = - 4 3 8 T -, etc. ; 

(5) sin A = |ff, tanA = ffjj, etc. 
in a right triangle, given : 

2. a = Vm 2 + n 2 , b = V2 mn ; calculate sin A. 

Vm 2 + n 2 



Ans. 



3. a = Vm 2 — mn, b = n ; calculate sec A. 



m + n 



m — n 



n 



/ — " * IVY) I yyi / y) 

4. a = Vm 2 +mw, c=m + n; calculate tan A. \ — — — -• 

2 2 

5. a = 2mn, b = m 2 — ?i 2 ; calculate cos A, —r~ — 

m 2 +n 2 

6. sinA^f, c- 200.5; calculate a. 120.3. 

7. cosA = .44, c = 30.5; calculated 13.42. 

8. tan A = Jf, 5 = ff ; calculate c. •& Vl30. 



20 PLANE TRIGONOMETRY, 

15. Functions of Complemental Angles. — In the rt. A ABC 

we have 

sin A = -, and cos B = -. (Art. 13.) 
c c 

.-. sin A = cosB. 

But B is the complement of A, since 
their sum is aright angle, or 90°; i.e., & 
B = 90°-A. 



Also, 




sin A 


= cosB 


= cos (90°- A) 


c 


cos A 


= sinB 


= sin (90°- A) 


c 


tan A 


= cotB 


= cot (90°- A) 


a 


cot A 


== tanB 


= tan (90°- A) 


b 

? 

a 


sec A 


= cosec B 


= cosec (90°- A) 


c 

~b' 


cosec A 


= secB 


= sec (90°- A) 


_ c 
a 



vers A = covers B = covers (90° — A) = 1 , 

c 

covers A = versB = vers (90°— A) =1 

c 

Therefore the sine, tangent, secant, and versed sine of an 
angle are equal respectively to the cosine, cotangent, cosecant, 
and coversed sine of the complement of the angle. 

16. Representation of the Trigonometric Functions by 
Straight Lines. — The Trigonometric functions were for- 
merly defined as being certain straight lines geometrically 
connected with the arc subtending the angle at the centre 
of a circle of given radius. 

Thus, let AP be the arc of a circle subtending the angle 
AOP at the centre. 



REPRESENTATION OF FUNCTIONS BY LINES. 21 

Draw the tangents AT, BT' meeting OP prodnced to T r , 
and draw PC, PD ± to OA, OB. 




Then PC was called the sine of the arc AP. 



oc 


U 


cosine 


a 


AT 


a 


tangent 


a 


BT' 


ii 


cotangent 


a 


OT 


ii 


secant 


a 


OT' 


a 


cosecant 


a 


AC 


a 


versed sine 


a 


BD 


ii 


coversed sine 


a 



Since any arc is the measure of the angle at the centre 
which the arc subtends (Art. 5), the above functions of the 
arc AP are also functions of the angle AOP. 

It should be noticed that the old functions of the arc above 
given, when divided by the radius of the circle, become the 
modern functions of the angle which the arc subtends at the 
centre. If, therefore, the radius be taken as unity, the old 
functions of the arc AP become the modern functions of the 
angle AOP. 

Thus, representing the arc AP, or the angle AOP by 6, we 
have, when A = OP = 1 ? 



22 PLANE TRIGONOMETRY. 

■"•-iihr-** 

and similarly for the other functions. 

Therefore, in a circle whose radius is unity, the Trigono- 
metric functions of an arc, or of the angle at the centre meas- 
ured by that arc, may be defined as follows : 

The sine is the perpendicular let fall from one extremity of 
the arc upon the diameter passing through the other extremity. 

The cosine is the distance from the centre of the circle to the 
foot of the sine. 

The tangent is the line which touches one extremity of the 
arc and is terminated by the diameter produced passing through 
the other extremity. 

The secant is the portion of the diameter produced through 
one extremity of the arc which is intercepted between the centre 
and the tangent at the other extremity. 

The versed sine is the part of the diameter intercepted 
between the beginning of the arc and the foot of the sine. 

Since the lines PD or OC, BT', OT', and BD are respect- 
ively the sine, tangent, secant, and versed sine of the arc 
BP, which (Art. 12) is the complement of AP, we see that 
the cosine, the cotangent, the cosecant, and the coversed sine of 
an arc are respectively the sine, the tangent, the secant, and the 
versed sine of its complement. 

EXAMPLES. 

1. Prove tan A sin A + cos A = sec A. 

2. " cot A cos A + sin A = cosec A. 

3. " (tan A - sin A) 2 + (1 - cos A) 2 = (sec A - l) 2 . 

4. " tan A + cot A = sec A cosec A. 

5. " (sin A + cos A) -f- (sec A + cosec A) = sin A cos A. 

6. « (1 + tanA) 2 + (1 + cot A) 2 = (sec A + cosec A) 2 . 



POSITIVE ASD NEGATIVE LINES. 



23 



7. Given tan A = cot 2 A ; find A. 

8. u sinA=cos3A; find A. 

9. - sin A = cos (45°- £A) ; find A. 

10. " tanA = cot6A; find A. 

11. " cot A = tan (45° + A) ; find A. 

17. Positive and Negative Lines. — Let AA' and BB' be 

two perpendicular right lines intersecting at the point 0. 
Then the position of an}' point in 
the line A A' or BB' will be deter- 
mined if we know the distance of 
the point from 0, and if we know 
also upon which side of the a'- 
point lies. It is therefore con- 
venient to employ the algebraic 
signs + and — , so that if dis- 
tances measured along the fixed 
line OA or OB from in one direction be considered 
positive, distances measured along OA' or OB' in the oppo- 
site direction from will be considered negative. 

This convention, as it is called, is extended to lines parallel 
to AA' and BB'; and it is customary to consider distances 
measured from BB' towards the right and from A A' upward* 
as positive, and consequently distances measured from BB' 
towards the left and from AA' downwards as negative. 



M 



O M 



B 



18. Trigonometric Functions of Angles of Any Magni- 
tude. — In the definitions of the trigonometric functions 
given in Art. 13 we considered only acute angles, i.e., angles 
in the first quadrant (Art. 5), since the angle was assumed 
to be one of the acute angles of a right triangle. TTe shall 
now show that these definitions apply to angles of any mag- 
nitude, and that the functions vary in sign according to the 
quadrant in which the angle happens to be. 



24 



PLANE TBIGONOMETEY. 



Let AOP be an angle of any mag- 
nitude formed by OP revolving from 
an initial position OA. Draw PM _L 
to AA'. Consider OP as always 
positive. Let the angle AOP be 
denoted by A ; then whatever be the 
magnitude of the angle A, the defini- 
tions of the trigonometric functions 
are 



p 




2nd 
Quad. 


Quadrant\P 




m\ 


^M 11. 


[m 

\ 3rd 

\Quad. 


0/ 


4th\ 


M' 




p 




Quad\ 


P 



. . MP » OM 

8 mA = — , cosA=— , 



tan A = 



A OP . A OM A 

secA = — — , cotA = , cosecA; 

OM MP 

I. When A lies in the 1st quadrant, 
MP is positive because measured from M 
upwards, OM is positive because measured 
from towards the right (Art. 17) ; and OP 
is positive. 

Hence in the first quadrant all the func- 
tions are positive. 

II. When A lies in the 2d quadrant, as 
the obtuse angle AOP, MP is positive 
because measured from M upwards, OM is 
negative because measured from towards 
the left (Art. 17), and OP is positive. 

Hence in the second quadrant 
MP 



B 

MP 
OM' 
OP 



MP 





sin A 



OP 



is positive ; 



cos A = — — is negative; 



tan A i 



MP 
: OM 



is negative; 



and therefore sec A and cot A are negative, and cosec A is 
positive (Art. 13). 



FUNCTIONS OF ANGLES. 



25 



III. When A lies in the 3d quadrant, 
as the reflex angle AOP, MP is negative 
because measured from M doivnwards, OM 
is negative, and OP is positive. 

Hence in the third quadrant the sine, 
cosine, secant, and cosecant, are negative, 
but the tangent and cotangent are positive. 

IV. When A lies in the 4th quadrant, 
as the reflex angle AOP, MP is negative, 
OM is positive, and OP is positive. 

Hence in the fourth quadrant the sine, 
tangent, cotangent, and cosecant are negative, 
but the cosine and secant are positive. 

The signs of the different functions are shown in the 
annexed table. 




Quadrant. 


I. 


II. 


III. 


IV. 


Sin and cosec 


+ 


+ 


- 


- 


Cos and sec 


+ 


- 


- 


+ 


Tan and cot 


+ 




+ 


- 



Note. — It is apparent from this table that the signs of all the functions in any 
quadrant are known when those of the sine and cosine are known. The tangent and 
cotangent are + or — , according as the sine and cosine have like or different signs. 

19. Changes in the Value of the Sine as the Angle in- 
creases from 0° to 360°. — Let A de- 
note the angle AOP described by the 
revolution of OP from its initial posi- 
tion OA through 360°. Then, PM 
being drawn perpendicular to AA f , 

• a MP 

sin A = , 

OP' 

whatever be the magnitude of the 
angle A. 




26 



PLANE TRIGONOMETRY. 



When the angle A is 0°. P coincides with A, and MP is 
zero ; therefore sin0° = 0. 

As A increases from 0° to 90°, MP increases from zero to 
OB or OP, and is positive; therefore sin 90° = 1. 

Hence in the 1st quadrant sin A is positive, and increases 
from to 1. 

As A increases from 90° to 180°, MP decreases from OP 
to zero, and is %)ositive; therefore sin 180° = 0. 

Hence in the 2d quadrant sin A is positive, and decreases 
from 1 to 0. 

As A increases from 180° to 270°, MP increases from 
zero to OP, and is negative; therefore sin 270° = — 1. 

Hence in the 3d quadrant sin A is negative, and decreases 
algebraically from to — 1. 

As A increases from 270° to 360°, MP decreases from OP 
to zero, and is negative; therefore sin 360°= 0. 

Hence in the 4th quadrant sin A is negative, and increases 
algebraically from — 1 to 0. 



20. Changes in the Cosine as the Angle increases from 0° 
to 360°. — In the figure of Art. 19 



cos A = 



OM 
OP* 



When the angle A is 0°, P coincides with A, and 
OM = OP ; therefore cos0°= 1. 

As A increases from 0° to 90°, OM decreases from OP to 
zero and is positive; therefore cos 90°= 0. 

Hence in the 1st quadrant cos A is positive, and decreases 
from 1 to 0. 

As A increases from 90° to 180°, OM increases from zero 
to OP, and is negative; therefore cos 180° =— 1. 

Hence in the 2d quadrant cos A is negative, and decreases 
algebraically from to — 1. 

As A increases from 180° to 270°, OM decreases from OP 
to zero, and is negative ; therefore cos 270° = 0. 



CHANGES IN THE TANGENT. 27 

Hence in the 3d quadrant cos A is negative, and increases 
algebraically from — 1 to 0. 

As A increases from 270° to 360°, OM increases from zero 
to OP, and is positive; therefore cos 360° =1. 

Hence in the 4th quadrant cos A is positive, and increases 
from to 1. 

21. Changes in the Tangent as the Angle increases 
from 0° to 360°. —In the figure of Art. 19 

tan A = - — 
OM 

When A is 0°, MP is zero, and OM = OP; therefore tan 
0°=0. 

As A increases from 0° to 90°, MP increases from zero to 
OP, and OM decreases from OP to zero, so that on both 
accounts tan A increases numerically; therefore tan90°=oo. 

Hence in the 1st quadrant tan A is positive, and increases 
from to oo. 

As A increases from 90° to 180°, MP decreases from OP 
to zero, and is positive, OM becomes negative and decreases 
algebraically from zero to — 1 ; therefore tan 180°= 0. 

Hence in the 2d quadrant tan A is negative, and increases 
algebraically from — oo to 0. 

When A passes into the 2d quadrant, and is only just greater than 
90°, tan A changes from +oo to — go. 

As A increases from 180° to 270°, MP increases from zero 
to OP, and is negative, OM decreases from OP to zero, and 
is negative; therefore tan 270°= oo. 

Hence in the 3d quadrant tan A is positive, and increases 
from to oo. 

As A increases from 270° to 360°, MP decreases from OP 
to zero, and is negative, OM increases from zero to OP, and 
is positive ; therefore tan 360° = 0. 

Hence in the 4th quadrant tan A is negative, and increases 
algebraically from — oo to 0. 



28 



PLANE TRIGONOMETRY. 



The student is recommended to trace in a manner similar 
to the above the changes in the other functions, i.e., the 
cotangent, secant, and cosecant, and to see that his results 
agree with those given in the following table. 

22. Table giving the Changes of the Trigonometric 
Functions in the Four Quadrants. 



Quadrant. 


I. 


II. 


III. 


IV. 


sin varies from 


+ 
to 1 


+ 
1 to 


to - 1 


- 1 to 


cos " ". 


+ 
1 toO 


to - 1 


- 1 toO 


+ 
to 1 


tan " " 


+ 
to GO 


— GO to 


+ 
to co 


— co to 


cot " " 


+ 

co toO 


to co 


+ 

co to 


to — co 


sec " " 


+ 

1 tO GO 


— GO tO — 1 


— 1 to — CO 


+ 

CO tO 1 


cosec " " 


+ 
00 to 1 


+ 

1 tO CO 


— CO tO — 1 


— 1 to — CO 


vers u " 


+ 

to 1 


+ 

1 to 2 


+ 

2 to 1 


+ 

1 toO 



Note 1. — The cosecant, secant, and cotangent of an angle A have the same sign 
as the sine, cosine, and tangent of A respectively. 

The sine and cosine vary from 1 to — 1, passing through the value 0. They are 
never greater than unity. 

The secant and cosecant vary from 1 to — 1, passing through the value co. They 
are never numerically less than unity. 

The tangent and cotangent are unlimited in value. They have all values from — qo 
to +oo. 

The versed sine and coversed sine vary from to 2, and are always positive. 

The trigonometric functions change sign in passing through the values and co, 
and through no other values. 

In the 1st quadrant the functions increase, and the cofunctions decrease. 

Note 2. — From the results given in the above table, it will be seen that, if the 
value of a trigonometric function be given, we cannot fix on one angle to which it 
belongs exclusively. 

Thus, if the given value of sin A be \, we know since sin A passes through all 
values from to 1 as A increases from 0° to 90°, that one value of A lies between <f 



RELATIONS BETWEEN FUNCTIONS. 



29 



and 90°. But since we also know that the value of sin A passes through all values 
between 1 and as A increases from 90° to 180°, it is evident that there is another 
value of A between 90° and 180° for which sin A = £. 



23. Relations between the Trigonometric Functions of 
the Same Angle. — Let the radius start 
from the initial position OA, and revolve FV 
in either direction, to the position OP. 
Let denote the angle traced out, and 
let the lengths of the sides PM, MO, 
OP be denoted by the letters a, b, a* 

The following relations are evident 
from the definitions (Art. 13) : 




cosec = 
I. tan (9 = 



1 a 1 

, sec# = 



sin 
sin 6 



COS0 



cot0 = 



tan 



cos# 



For 



, * a c sin 6 

tan = - = -.== - 



6 

c 



COS0 



II. sin 2 + cos 2 <9 = l. 



Forsin 2 + cos 2 = ^ + ^ = ^^ = l. 
c 2 c 2 c 2 



III. sec 2 <9=l + tan 2 <9. 



For sec 2 0=-: 



b 2 + a 2 



b 2 b 2 
IV. cosec 2 0=:l + cot 2 0. 



! + ^=l+tan*A 
b 1 



For cosec 2 (9=-'=^±^ 2 =l+-=l+cot 2 ft 




cot 



Formulae L, II., III., IV. are very important, and must be 
remembered. 



* a, by c are numbers, being the number of times the lengths of the sides contain 
some chosen unit of length. 



30 PLANE TRIGONOMETRY. 

24. Use of the Preceding Formulae. 

I. To express all the other functions in terms of the sine. 



Since sin 2 + cos 2 = 1, . \ cos = ± V 1 — sin 2 0. 
sin sin 



tan0 = 



cos VI -sin 2 



,. n 1 L Vl-sin 2 

cot0 = - = ± — — — - 

tan sin 

sec0 = 
cosec0 = 



cos (9 Vl-sin 2 

1 



sin0 

II. To express all the other functions in terms of the tan- 
gent. 

Since tan0 = ^, 

COS0 

e - zj i^/)^/] tan0 . tan 
sin = tan cos = = ± — 



COS0: 



sec0 Vl+tan 2 

1 1 



sec0 " Vl + tan 2 
1 



cot0 = — — seed = ± Vl + tan*0. 
tan 



a 1 , Vl + tan 2 <9 

cosec = ~ — - = ± - — - — 

sin tan 

Similarly, any one of the functions of an angle may be 
expressed in terms of any other function of that angle. 
The sign of the radical will in all cases depend upon the 
quadrant in which the angle lies. 

25. Graphic Method of finding All the Functions in Terms 
of One of them. ^B 

To express all the other functions in 
terms of the cosecant. 

Construct aright triangle ABC, hav- 
ing the side BC = 1. Then a V coseca-i " 




RELATIONS BETWEEN FUNCTIONS. 



°1 



A AB AB A -p 

cosec A = = — = AB. 

BC 1 



Now 



\ AC = 

sin A = 


= ±Vc( 
BC 
AB 

AB 


3sec 2 A — 1. 

1 
cosec A' 




cosA = 
tanA = 


Vcosec 2 A - 
cosec A 

+ 1 


-i 



-A-C Vcosec 2 A — 1 
and similarly the other functions may be expressed in terms 
of cosec A.. 

26. To find the Trigonometric Functions of 45°. — Let 
ABC be an isosceles right triangle in which 
CA = CB. 
Then CAB = CBA = 45°. 
Let AC = m = CB. £y 

Then 

AB 2 = AC 2 + CB 2 = m 2 + m 2 = 2m 2 . 
.-. AB = mV2. 

BC = m _ 1 
AB m V2 
AC^_m_ 
AB 




sin45° = — = 



cos 45°: 



V2 
1 



to 



V2 



V2 

cot 45° 
m 

sec 45° = V2. cosec 45° = V2. 



tan 45° =^=™ = 1. 
AC m 



= 1. 



27. To find the Trigonometric Func- 
tions of 60° and 30°. —Let AB be an 

equilateral triangle. Draw AD perpen- 
dicular to BC. Then AD bisects the 
angle BAC and the side BC. Therefore 
BAD = 30°, and ABD = 60°. 




32 



Let 
Then 



PLANE TRIGONOMETRY. 
BA = 2m. ,\ BD = m. 

AD = V4m 2 — m 2 = m V3. 



.\ sin 60° = 



AD := mV5 
AB : 2m 



= ±V3. 



cosec60°=- 



"V3 



cn o BD 1 

cos 60 = — = — 
BA 2 



.-. sec 60° = 2. 



tan60°=^ = 2?LV3 = V3. 



BD 



m 



cot 60° = 



vers 60°= 1 - cos 60°= 1 - - = 1. 

2 2 



V3 



Also sin30» = ?5 = -*»- = !. 

AB 2m 2 



.-. cosec30°=2. 



cos 30°: 



,AD_ mV3 
: AB 



tan 30° =5*5 = 



2m 
m 



1 



DA 



mV3 V3 

EXAMPLES. 



. S ec 30° = —. 
V3 

cot30°=V3. 



1. Given sin0 = -; find the other 
5 

trigonometric functions. 

Let BAC be the angle, and BC be 
perpendicular to AC. Represent BC 
by 3, AB by 5, and consequently AC £ 
by V25-9 = 4. 

Then CO s0 = A§4 

AB 5 

a AB 5 
SeC * = AC = 4' 

a AB 5 
cosec0 = - = -. 



>iB 



REDUCTION OF FUNCTIONS. 33 

o 3 5 

2. Given sin = - ; find tan 6 and cosec ft Ans. -> -• 

5 4 o 

3. Given cos0 = -; find sin 6 and cot ft fV2, — — • 

o 2\2 

1 Vl5 



4. Given sec# = 4; find cot 6 and sin 0. 



Vl5 



5. Given tan = V3 ; find sin 6 and cos ft \ V3, -• 

12 5 

6. Given sin 6 = — ; find cos 0. — • 

13 -L^ 



7. Given cosec 6 = 5 ; find sec0 and tan ft 



1 



2V6 2V6 



— ; find sin 6 and cot ft — ? — . 

9 41 40 



8. Given sec = — ; find sin and cot ft 



2 a/5 3 

9. Given cot = — - : find sin0 and sec ft — -> -• 

V5 3 2 

3 
10. Given sin0 = - ; find cos0, tan0, and cot ft 

4 V7 3V7 V7 



4 7 3 



11. Given sin = - ; find tan ft 



c Vc 2 - & 2 

-*o /-»• a 2mn n * j. a 2mn 

12. Given sin0 = — ; ; find tan ft — -• 

rnr+nr mr—n* 

28. Reduction of Trigonometric Functions to the 1st 
Quadrant. — All mathematical tables give the trigonometric 
functions of angles between 0° and 90° only, but in practice 
we constantly have to deal with angles greater than 90°. 
The object of the following six Articles is to show that the 
trigonometric functions of any angle, positive or negative, 
can be expressed in terms of the trigonometric functions of 
an angle less than 90°, so that, if a given angle is greater 
than 90°, we can find an angle in the 1st quadrant whose 
trigonometric function has the same absolute value. 



34 



PLANE TRIGONOMETRY. 



29. Functions of Complemental Angles. — Let AA' ? BB' 

be two diameters of a circle at right angles, and let OP and 
OP' be the positions of the radius for b 

any angle AOP = A, and its comple- 
ment AOP'= 90°- A (Art. 12). 

Draw PM and P'M' at right angles 
toOA. 

Angle OP'M'= BOP'= AOP = A. 

Also OP=OP'. 

Hence the triangles OPM and OP'M' are equal in all 

respects. 

.-. P'M'=OM. ., ™' = om 
OP' OP 

.-. sin (90° — A) = cos AOP = cos A. 

OM' = PM 
"■ OP' _ op' 

cos (90° - A) = sin AOP = sin A. 




Also, 



OM' = PM. 



Similarly, tan (90°- A) = tanAOP'= 



P'M' OM 



= cot A. 



OP' OP 

The other relations are obtained by inverting the above. 

30. Functions of Supplemental An- 
gles. — Let OP and OP' be the positions 
of the radius for any angle AOP = A, p 
and its supplement AOP'=180°— A , 
(Art. 12). 

Since OP = OP', and POA = P'OA', 
the triangles POM and P'OM' are 
geometrically equal. 

P'M ' PM 
.-. sin(180°-A) = sinAOP'=^ r =^ = sinA, 




cos (180°- A) = cos AOP'= 



OM' 
OP' 



-OM 
OP 



= — cos A, 



REDUCTION OF FUNCTIONS. 



35 



tan (180°- A) = tan AOP'= =^L = f^ 1 = - tan A. 

Similarly the other relations may be obtained. 



31. To prove sin (90°+ A) = cos A, cos (90°+ A) = — sin A, 
and tan (90°+ A) = - cot A. 

Let OP and OP' be the positions 
of the radius for any angle AOP = A, 
and AOP'=90°+A. 

Since OP = OP', and AOP = P'OB 
= OP'M', the triangles POM and 
P'OM' are equal in all respects. 

.-. sin(90 o +A) = sinAOP'=^ = || = cosA, 




cos (90°+ A) = cos AOP'== 
tan (90°+ A) = tan AOP' = 



OM' 



OP' 
P'M' 



-PM 
OP 

OM 



: — sin A, 



OM' - PM 



= — cot A. 



32. Toprovesin(180°+A) = -sinA, 
cos (180°+ A) = - cos A, and tan (180° 
+ A) = tan A. 

Let the angle AOP = A; then the 
angle AOP', measured in the positive 
direction, = (180°+ A). 

The triangles POM and P'OM' are 
equal. 

P'M' 




sin (180°+ A) = sin AOP': 



PM 



OP' 



OP 



= — sin A, 



cos (180°+ A) =cos AOP' = §^ == =^ = - cos A, 



OP' 



OP 



tan (180°+ A) = tan AOP' = — = — ™ = tan A. 
} OM' - OM 



36 



PLANE TRIGONOMETRY. 



33. To prove sin ( — A) = — sin A, cos ( — A) = cos A, 

tan ( — A) = — tan A. B 

Let OP and OP' be the positions of 
the radius for any equal angles AOP 
and AOP' measured from the initial 
line AO in opposite directions. Then 
if the angle AOP be denoted by A, the 
numerically equal angle AOP' will be 
denoted by —A (Art. 4). 

The triangles POM and P'OM are geometrically equal. 




P'M 



.\ sin(-A) = sinAOP' = ^^ = 



PM 



OP' 



OP 



sin A, 



cos ( — A) = cos AOP' : 



OM OM A 

= = cos A, 

OP' OP 



tan (- A) = tan AOP': 



P'M 



PM 



OM OM 



= — tan A. 



34. To prove sin (270° + A) = sin (270° - A) = 
and cos (270° + A) = - cos (270°- A) B 

= sin A. 

Let the angle AOP = A ; then the 
angles AOQ and AOE, measured in 
the positive direction, =(270°— A) A 
and (270°+ A) respectively. 

The triangles POM, QON, and EOL 
are geometrically equal. 



— cos A, 




EL = QN = OM. 



EL 
OE = 



. QN _ - OM 



OQ 



OP 



sin (270°+ A) = sin (270°- A) = - cos A. 



Also, 



OL 
OE : 



ON 



OQ 



PM 
OP' 



cos (270° + A) = - cos (270°- A) = sin A. 




VALUES OF THE FUNCTIONS. 37 

35. Table giving the Values of the Functions of Any Angle 
in Terms of the Functions of an Angle less than 90°. — The 

foregoing results, and other similar ones, which may be 
proved in the same manner, are here collected for reference. 

Quadrant II. 

sin (180°- A) = sin A. sin (90°+ A) = cos A. 

cos (180° - A) = - cos A. cos (90° + A) = - sin A. 

tan (180°- A) = - tan A. tan (90°+ A) = — cot A. 

cot (180°- A)==- cot A. cot (90°+ A) = - tan A. 

sec (180°- A) = - sec A. sec (90°+ A) = - cosec A. 

cosec (180°— A) = cosec A. cosec (90°+ A) == sec A. 

Quadrant III. 

sin (180° + A) = - sin A. sin (270° - A) = - cos A. 

cos (180°+ A) = - cos A. cos (270°- A) == - sin A. 

tan (180°+ A) = tan A. tan (270°- A) = cot A. 

cot (180° + A) = cot A. cot (270° - A) = tan A. 

sec (180°+ A) = - sec A. sec (270°- A) = - cosec A. 

cosec (180° + A) = — cosec A. cosec (270° — A) = — sec A. 

Quadrant IV. 

sin (360°- A) = - sin A. sin (270°+ A) = - cos A. 

cos (360°- A) = cos A. cos (270°+ A) = sin A. 

tan (360°- A) = - tanA. tan (270°+ A) = - cot A. 

cot (360°- A) = - cot A. cot (270°+ A) = - tan A. 

sec (360°- A) = sec A. sec (270°+ A) = cosec A. 

cosec (360°— A) = — cosec A. cosec (270°+ A) = — sec. A. 

Note. — These relations* may be remembered by noting the following rules : 

When A is associated with an even multiple of 90°, any function of the angle is 
numerically equal to the same function of A. 

When A is associated with an odd multiple of 90°, any function of the angle is 
numerically equal to the corresponding co function of the angle A. 

The sign to be prefixed will depend upon the quadrant to which the angle belongs 
(Art. 5), regarding A as an acute angle. 

♦Although these relations have been proved only in case of A, an acute angle, 
they are true whatever A may be. 



38 



PLANE TRIGONOMETRY. 



Thus, cos (270° -A) = — sin A; the angle 270° — A being in the 3d quadrant, and 
its cosine negative in consequence. 
For an angle in the 

First quadrant all the functions are positive. 
Second quadrant all are negative except the sine and cosecant. 
Third quadrant all are negative except the tangent and cotangent. 
Fourth quadrant all are negative except the cosine and secant. 

36. Periodicity of the Trigonometric Functions. — Let 

AOP be an angle of any magnitude, as in the figure of 
Art. 18 ; then if OP revolve in the positive or the negative 
direction through an angle of 360°, it will return to the 
position from which it started. Hence it is clear from the 
definitions that the trigonometric functions remain un- 
changed when the angle is increased or diminished by 360°, 
or any multiple of 360°. Thus the functions of the angle 
400° are the same both in numerical value and in algebraic 
sign as the functions of the angle of 400°— 360°, i.e., of the 
angle of 40°. Also the functions of 360°+ A are the same 
in numerical value and in sign as those of A. 

In general, if n denote any integer, either positive or 
negative, the functions ofnx 360°+ A are the same as those 
of A. 

Thus the functions of 1470°= the functions of 30°. 

If 6 denotes any angle in circular measure, the functions 
of (2 mr + 6) are the same as those of 0. Thus 

sin (2 71-n- + 6) = sin 0, cos (2 ?i7r + 0) = cos 0, etc. 

By this proposition we can reduce an angle of any magni- 
tude to an angle less than 360° without changing the values 
of the functions. It is therefore unnecessary to consider 
the functions of angles greater than 360° ; the formulae 
already established are true for angles of any magnitude 
whatever. 

EXAMPLES. 

Express sin 700° in terms of the functions of an acute 
angle. 

sin 700°= sin (360°+ 340°) = sin 340°= sin (180°+ 160°) 
= sinl60 o =-sin20 o . 



EXAMPLES. 39 

Express the following functions in terms of the functions 
of acute angles : 

1. sin 204°, sin 510°. Ans. -sin 24°, sin 30°. 

2. cos (-800°), cos 359°. cos 80°, cosl°. 

3. tan 500°, tan 300°. -tan 40°, -cot 30°. 

Find the value of the sine, cosine, and tangent of the 
following angles : 

4. 150°. Ans. 1, -iV3, — . 

2' V3 

5. -240°. |V3, -% -V3. 

6. 330°. -1, *V3, --1-. 

2 V3 

7. 225°. — , — ^ 1. 

V2 V2 

Find the values of the following functions : 

8. sin 810°, sin (-240°), cos 210°. Ans. 1, fV3, -|V& 

9. tan (-120°), cot 420°, cot 510°. V3, — , 1. 

V3 

10. sin 930°, tan 6420°. --, — . 

2 V3 

11. cot 1035°, cosec570°. -1, -2. 

37. Angles corresponding to Given Functions. — When an 
angle is given, we can find its trigonometric functions, as in 
Lrts. 26 and 27 ; and to each value of the angle there is 
3ut one value of each of the functions. But in the converse 
Droposition — being given the value of the trigonometric 
functions, to find the corresponding angles — we have seen 
(Art. 36) that there are many angles of different magnitude 
which have the same functions. 

If two such angles are in the same quadrant, they are 
represented geometrically by the same position of OP, so 
that they differ by some multiple of four right angles. 



40 



PLANE TRIGON02IJETRY. 



p 










P 


/ 




N ^^ 




y 







M 



If we are given the value of the sine of an angle, it is 
important to be able to find all the angles which have that 
value for their sine. 

38. General Expression for All Angles which have a 
Given Sine a. — Let be the centre of the unit circle. 
Draw the diameters AA', BB', at right b 

angles. 

From draw on OB a line ON, so 
that its measure is a. 

Through 1ST draw PP' parallel to 
AA'. Join OP, OP', and draw PM, 
P'M', perpendicular to AA'. 

Then since MP = M'P' = ON = a, 
the sine of AOP is equal to the sine of AOP'. 

Hence the angles AOP and AOP' are supplemental (Art. 
30), and if AOP be denoted by a 9 AOP' will = it - <*. 

Now it is clear from the figure that the only positive 
angles which have the sine equal to a are a and it — a, and 
the angles formed by adding any multiple of four right 
angles to a and tt — a. Hence, if 6 be the general value of 
the required angle, we have 

6 = 2mr + a, or 6 = 2nir + 7r — a, .... (1) 

where n is zero or any positive integer. 

Also the only negative angles which have the sine equal 
to a are — (tt + o), and — (2tt— a), and the angles formed 
by adding to these any multiple of four right angles taken 
negatively ; that is, we have 

0=2n7T-(TT + a), 6 = 2mT-(27T-a), . . (2) 

where n is zero or any negative integer. 

Now the angles in (1) and (2) may be arranged thus : 

2ri7T + a, (2n + l)7T — a, (2n — 1)tt — a, (2?i — 2)ir + a, 

all of which, and no others, are included in the formula 

e = n7T + (-l) n a, (3) 



GENERAL EXPRESSION FOR ANGLES. 



41 



where n is zero, or any positive or negative integer. There- 
fore (3) is the general expression for all angles which have 
a given sine. 

Note. — The same formula determines all the angles which have the same 
cosecant as a. 

39. An Expression for All Angles with a Given Cosine a. — 

Let be the centre of the unit circle. 
Draw A A', BB', at right angles. 

From draw OM, so that its meas- 
ure is a. 

Through M draw PP' parallel to 
BB'. Join OP, OP'. 

Then since OM = a, the cosine of 
AOP is equal to the cosine of AOP'. 

Hence, if AOP = a, AOP'= - a. 

Now it is clear that the only angles which have the cosine 
equal to a are a and — a, and the angles which differ from 
either by a multiple of four right angles. 

Hence if 6 be the general value of all angles whose cosine 
is a, we have 

= 2 utt ± a, 

where n is zero, or any positive or negative integer. 

Note. — The same formula determines all the angles which have the same secant 
or the same versed sine as a. 




40. An Expression for All Angles with a Given Tangent a. 

— Let be the centre of the unit 
circle. Draw AT, touching the circle 
at A, and take AT so that its measure 
is a. 

Join OT, cutting the circle at ~PfiC- 
and P'\ 

Then it is clear from the figure that 
the only angles which have the tan- 
gent equal to a are a and it + a, and the angles which differ 




42 PLANE TRIGONOMETRY. 

from either by a multiple of four right angles. Hence if 6 
be the general value of the required angle, we have 

6 — 2mr-{-a, and 2ft7r + 7r+a (1) 

Also, the only negative angles which have the tangent 
equal to a are — (?r — a), and — (2?r — a), and the angles 
which differ from either by a multiple of four right angles 
taken negatively ; that is, we have 

= 27nr-(7r-a), and 2nir - (2tt - a), . . (2) 

where n is zero or any negative integer. 

Xow the angles in (1) and (2) may be arranged thus : 

27iir + a, (2n + l)7r + a, (2n — l)ir + a, (2n — 2) ?r + a, 

all of which, and no others, are included in the formula 

= 7iw + a, (3) 

where n is zero, or any positive or negative integer. There- 
fore (3) is the general expression for all angles which have 
a given tangent. 

Note. — The same formula determines all the angles which have the same cotan- 
gent as a. 

EXAMPLES. 

1. Find six angles between — 4 right angles and + 8 
right angles which satisfy the equation sinA = sin 18°. 
We have from (3) of Art. 38, 

= W7r + (_l)»JL or A = nx 180°+ (-1)»18°. 

Put for n the values — 2, — 1, 0, 1, 2, 3, successively, 
and we get A = - 360° + 18°, - 180° - 18°, 18°, 180° - 18°, 
360°+ 18°, 540°- 18°; 

that is, - 342°, - 198°, 18°, 162°, 378°, 522°. 

Note. — The student should draw a figure in the above example, and in each 
example of this kind which he works. 



TRIGONOMETRIC IDENTITIES. 43 



2. Find the four smallest angles which satisfy the 

1 



equations (1) sinA = -, (2) sinA = — -, (3) sinA = - T -? 



(4) sinA 

v J 2 Ans. (1) 30°, 150°, - 210°, - 330 

(2) 45°, 135°, - 225°, - 315 

(3) 60°, 120°, - 240°, - 300' 

(4) _ 30°, - 150°, 210°, 330°. 

41. Trigonometric Identities. — A trigonometric identity 
is an expression which states in the form of an equation a 
relation which is true for all values of the angle involved. 
Thus, the relations of Arts. 13 and 23, and all others that 
may be deduced from them by the aid of the ordinary 
formulae of Algebra, are universally true, and are therefore 
called identities; but such relations as sin# = |-, cos 8 = J, 
are not identities. 

EXAMPLES. 

1. Prove that sec — tan • sin 6 = cos 6. 

Here sec 6 - tan<9 sin 6 = — ^2^ sin (Art. 24) 



sec#- 


, n . A ' 1 sin# . A 

- tan0 sin 6 = sin 

cos 6 cos# 




l-sin 2 <9 




COS0 




_ COS 2 



(Art. 24) 
cos<9 v } 

— cos 8. 

2. Prove that cot 6 — sec 6 cosec 6 (1 — 2 sin 2 0) = tan 0. 

cot 6 — sec cosec 9 (1 — 2 sin 2 6) 

cos 6 1 1 



sin cos sin 

cos 2 *9-l + 2sin 2 (9 
sin cos 



(1-2 sin 2 6) (Art. 24) 



44 PLANE TRIGONOMETRY. 



cos 2 6 -(sin 2 6 + cos 2 6) +2 sin 2 6 , A , . N 

= * r- J ^-^ ■ (Art. 24) 

sin cos 

sin 2 sin 6 , A 

= tan 0. 



sin cos cos# 

Note. — It will be observed that in solving these examples we first express the 
other functions in terms of the sine and cosine, and in most cases the beginner will 
find this the simplest course. It is generally advisable to begin with the most com- 
plicated side and work towards the other. 

Prove the following identities : 

3. cos tan 6 = sin 0. 

4. cos 6 = sin 6 cot 0. 

5. (tan + cot 0) sin cos = 1. 

6. (tan — cot 0) sin cos = sin 2 — cos 2 0. 

7. sin 2 -r- cosec 2 = sin 4 0. 

8. sec 4 — tan 4 = sec 2 + tan 2 0. 

9. (sin0-cos0) 2 =l-2sin0cos0. 

10. 1 - tan 4 = 2 sec 2 - sec 4 (9. 

11. 1 + cos g = (cosec fl + cot fl) 2 . 
I_cos0 V ; 

12. (sin + cos 0) 2 + (sin0 - cos 6>) 2 = 2. 

13. sin 4 - cos 4 = sin 2 - cos 2 (9. 

14. sin 2 + vers 2 = 2(l-cos0). 

15. cot 2 0-cos 2 = cot 2 0cos 2 0. 

EXAMPLES. 

In a right triangle ABC (see figure of Art. 15) given : 
1. a=p 2 -\-pq, c = q 2 +pq-, calculate cot A, 



p 

2. b = Zm -7- n, c = In -=- m ; calculate cosec A. 



Vn 4 



EXAMPLES. 45 

3. sin A == -, c = 20.5 ; calculate a. Ans. 12.3. 



4. 


Given cot -J- A = tan A; find A. 




5. 


tt 


sin A = cos2A; find A. 




6. 


tt 


cot A = tan 6 A ; find A. 




7. 


tt 


tan A = cot 8 A ; find A. 




8. 


tt 


sin2A = cos3A; find A. 




9. 


a 


o 
sin A = - : find cos A and tan A. 
3' 


V5 2 
3' V5 


10. 


it 


cos A =-; find sin A and tan A 
5 


3 3 
5' I 


11. 


a 


cosec A = - ; find cos A and tan A. 
3' 


V7 3 

4'V7 


12. 


(( 


sin A = : find cos A and tan A. 

V3 


(2 1 

W V 2 


13. 


a 


cos A = b ; find tan A and cosec A. 





vr 



14. " sin A =.6: find cos A and cot A. 



b Vl - b* 

4 4 



5 J 3 



15. " tan A =-; find sin A. 



5 V41 

16. " cotA = — ; find sec A and sin A. — , — 

15' 17' 17 

17. " sin A = — : find cos A. — 

13' 13 

18. " cos A = .28; find sin A. .96. 

19. " tan A =-;findsinA. ~. 

3 5 

20. " sin A =i; find cos A. |V2. 

o 



46 



PLANE TRIGONOMETRY. 



21. Given tan A = -; find sin A and sec A. Ans. -, -• 
o o o 



22. 



23. 



" tan 6 = -; find sin0 and cos0. 
b 



a 



" cos0 = -; find sin and cot 6, 
a 



Va 2 + V Va 2 + b 2 
1 



Va 2 



Va 2 -1 



24. If sin = a, and tan = b, prove that 
(l-a 2 )(l + 6 2 ) = l. 

Express the following functions in terms of the functions 
of acute angles less than 45° : 



25. sinl68°, sin 210°. 

26. tan 125°, tan 310°. 

27. sec 244°, cosec281°. 

28. sec930° ? cosec (- 600°). 

29. cot460°, sec 299°. 

30. tan 1400°, cot (-1400°). 



Ans. sin 12°, -sin 30°. 

-cot35° ? -cot 40°. 

-cosec 26°, -sec 11°. 

-sec30° ? sec 30° 

-tan 10°, cosec 29°. 

-tan40° ? cot 40° 



Find the values of the following functions : 
31. sin 120°, sin 135°, sin 240°. Ans, ' 



32. cos 135°, tan 300°, cosec 300°. 

33. sec 315°, cot 330°, tan 780°. 

34. sin480° ? sin495° ? sin 870°. 



35. tan 1020°, sec 1395°, sin 1485° 



V3 



2 


V2 


2 


1 
V2' 


-V3, - 


2 
*V5 


V2, 


-V3, 


V3. 




V3 1 

2 ' V2 


1 
t 2 



_V3, V2, * 

V2 



EXAMPLES. 47 

36. sin (-240°), cot (-675°), cosec (- 690°). 

Ans. 2@, 1, 2 . 

37. cos (-300°), cot (-315°), cosec (- 1740°). 

2' X ' V3 

38. tan 3 660°, cos 3 1020°. -3V3, -• 

8 

Find the value of the sine, cosine, and tangent of the 
following angles : 

39. -300°. Ans. — - V3 

2 ' 2' 

40. -135°. K, -—, 1. 

V2 V2 

1 V3 1 

2' 2' V3" 

42. -840°. -^, -| V3. 

43. 1020°. -^, | -V3. 

44. (2 M + 1)^-|. ^?, -| -V3. 



41. 750°. 



45. (2n-lW + -- -i -^ — • 

V ' T 6 2' 2 ' V3 

Prove, drawing a separate figure in each case, that 

46. sin340°=sin(-160°). 

47. sin (-40°) = sin 220°. 

48. cos320°=-cos(140°). 

49. cos ( - 380°) = - cos 560°. 

50. cos 195° = - cos ( - 15°) . 

51. cos380°=-cos560°. 



48 PLANE TRIGONOMETRY. 

52. cos ( - 225°) = - cos ( - 45°) . 

53. cos 1005°=- cos 1185°. 

54. Draw an angle whose sine is — 

& 2 

55. " " " " cosecant is 2. 

56. " « " " tangent is 2. 

57. Can an angle be drawn whose tangent is 427 ? 

58. " « " " « " cosine is-? 

4 

59. " " " " " " secant is 7 ? 

60. Find four angles between zero and + 8 right angles 
which satisfy the equations 

(1) sin A = sin 20°, (2) sin A = ~> (3) sin A = --• 

Ana. (1) 20°, 160°, 380°, 520°; 

,r>\ 57T 1 IT 137T 15tT 

w T T' T' ~4~ ; 

• r>\ 87T 137T 227T 277T 

(3) y, — , — , — • 

61. State the sign of the sine, cosine, and tangent of each 
of the following angles : 

(1) 275°; (2) -91°; (3) -193°; (4) -350°; 
(5) -1000°; (6) 2nw + — ■ 

Ans. (1) -,+,-; (2) -,-,+; (3) +,-,-; 
(4) +,+,+; (6) +,+,+; (6) +, -, -. 

Prove the following identities : 

62. (sin 2 + cos 2 0) 2 =l. 

63. (sin 1 * - 008*0)*= 1 - 4cos 2 + 4cos 4 0. 

64. (sin 6 + cos 0) 2 = 1 + 2sin 6 cos 6. 
66. (sec 6 - tan 6) (sec 6 + tan 6) = 1. 






EXAMPLES. 49 

66. (cosec 6 — cot 0) (cosec + cot 0) = 1. 

67. sin 3 + cos 3 = (sin (9 + cos (9) (1 - sin cos 0) . 

68. sin 6 6 + cos 6 (9 = sin 4 + cos 4 - sin 2 (9 cos 2 (9. 

69. sin 2 tan 2 6 + cos 2 (9 cot 2 = tan 2 + cot 2 0-1. 

70. sin0 tan 2 0+ cosec sec 2 — 2 tan sec0 = cosec 0— sin0. 

71. cos 3 - sin 3 == (cos - sin0) (1 + sin cos0). 

72. sin 6 + cos 6 = 1 - 3 sin 2 cos 2 0. 

73. tana + tan/}= tana tan/? (cota + cot/3). 

74. cot a + tan /? = cot a tan /? (tan a + cot /?). 

75. 1 — sin a = (1 + sin a) (sec a — tan a) . 2 

76. 1 + cos a = (1 — cos a) (cosec a + cot a) ? 

77. (1 + sina + cosa) 2 = 2(1 + sina) (1 + cos a). 

78. (1 — sina — cosa) 2 (l + sina+cosa) 2 = 4sin 2 acos 2 a, 

79. 2 vers a — vers 2 a = sin 2 a. 

80. vers a (1 + cos a) = sin 2 a. 



50 



PLANE TRIGONOMETRY. 



CHAPTER III. 



TKIG0N0METEI0 FUNCTIONS OF TWO ANGLES. 

42. Fundamental Formulae. — We now proceed to express 
the trigonometric functions of the sum and difference of 
two angles in terms of the trigonometric functions of the 
angles themselves. 

The fundamental formulae first to be established are the 
following : 



sin (x + y) = sin x cos y + cos x siny 
cos {x + y) = cos x cosy — sin#sin^ 
sin (x — y) = sin x cos y — cos x sin y 
cos (x — y) = cos x cosy + sin # sin y 



(i) 

(2) 
(3) 
(4) 



Note. — Here x and y are angles; so that (x + y) and (x — y) are also angles. 
Hence, sin (x + y) is the sine of an angle, and is not the same as sin x + sin y. 
Sin (x + y) is a single fraction. 
Sin x + sin y is the sum of two fractions. 

43. To prove that 

sin {x + y) = sin x cos y + cos x sin y, 
and cos (x + y) = cos x cos 2/ — sin x sin ?/. 

Let the angle AOB = x, and the 
angle BOC = y ; then the angle 
AOC = x + y. 

In OC, the bounding line of the 
angle (x + y), take any point P, 
and draw PD, PE, perpendicular 
to A and OB ? respectively ; draw 
EH, EK, perpendicular to PD and OA. 




FUNDAMENTAL FORMULA. 



51 



Then angle EPH = 90°- HEP = HEO = AOE = x. 

- z , x DP EK + PH EK , PH 
Bm(x + y) = — = 0p =oF + OP 

EK OE* PH PE 

" OE " OP PE " OP 



= sin x cos y + cos x sin y. 
HE 



, , N OB OK 
008(^ + 2/) = — = -^ 



OK 
OP 



HE 
OP 



OK OE 

OE ' OP 



HE PE 

PE ' OP 



= cos x cos y — sin x sin y . 

Note. — These two formulae have been obtained by a construction in which x + y 
is an acute angle; but the proof is perfectly general, and applies to angles of any 
magnitude whatever, by paying due regard to the algebraic signs. For example, 

Let AOB = #, as before, and BOC = t/; 
then AOC, measured in the positive direc- 
tion, is the angle x + y. 

In OC, take any point P, and draw PD, 
PE, perpendicular to OA and OB produced; 
draw EH and EK perpendicular to PD and 
OA. 

Then, angle EOK = 180° - x; 



and 



sin (x + y) = 



EPH = EOK = 180° -x; 
COE = y - 180°. 

DP EK - PH 



OP 



OP 




EK OE + PH t PE 
OE OP PE * OP 

= - sin (180° - x) cos (y - 180°) + cos (180° - x) sin (y - 180°) 

= sin x cos y + cos x sin y. (Art. 35) 

Q os(x + y)=^ = OK ± ^R = OK + m 
" OP OP OP OP 

= OK OE + EH PE 
OE *OP PE ' OP 



* The introduced line OE is the only line in the figure which is at once a side of 
two right triangles (OEK and OEP) into which EK and OP enter. A similar 
remark applies to PE. 



52 



PLANE TRIGONOMETRW 



m cos (180° - x) cos (y - 181K) + sin (180° - x) sin (y - 180°) 
= cos x cos y — sin 25 sin y. ( Art. 35) 

The student should notice that the words of the two proofs are very nearly the 

■ame, 

44. To prove that 

sin (x — //) = sin x cos?/ — ooscc sin?/, 
and cos (x — //) — cos a* COB // + sin x sin ?/. 

Let the angle AOB be denoted 
by x } and COB by //; then the angle 
AOC = 0-y. 

In 00 take any point P*, and 
draw PI), PE, perpendicular to 
OA and OB respectively ; draw 
KH, E.K, perpendicular to PD and 
OA respectively. 




Then the angle EPH = 90* - HEP = BEH= AOB = x. 

nr 
OP 



V 9) Qp 



or 



OP 



= EK OE HP PE 

= oe' or pe'op 

bs sin X 0O8Jf — 008 8 sin//. 

^-a» n OD OK + 1H OK , EH 

cos (a; — ?/) = — = as r- 

v •" OP or op op 
_OK OE EH PE 

ss oe'op pe'op 

bb COS X cox y + sin.r sin//. 

Notk 1. — The sign in the expression of the sine hi the MMM as it is in the angle 

expanded; In the ooeine it i« the opposite 



* P is taken in the line bounding the engle nnder oonelderatton; i.e.. \oc. 



SINE AND COSINE. 



53 



Note 2. — In this proof the angle x — y is acute ; but the proof, like the one given 
in Art. 43, applies to angles of any magnitude whatever. For example, 

Let AOB, measured in the positive \ n 

direction, = x, and BOC = 2/. Then \ 

AOC = *-y. Hp— ^E 

In OC take any point P, and draw C- 
PD, PE, perpendicular to OA and OB 
produced: draw EH, EK, perpendicu- 
lar to DP and AO produced. 
Then, 

angle EPH = EOK = AOB = 360° - x, 

and POE = 180°-2/. 




«n (x — y) 



coB(x-y) 



_PD_ EK-HP 
OP OP 

= EK OE HP PE 
OE *OP PE ' OP 

= sin (360° - x) cos (180° - y) — cos (360° - x) sin (180° - y) 

= (— sin x) ( — cos y) — cos x sin y 

= sin x cos y — cos x sin y. 

OK + HE 



OD_ 
OP" 



OP 



= _OK OE_HE PE 
OE ' OP PE ' OP 

= - cos (360° - x) cos (180° - y) - sin (360° - x) sin (180° - y) 

= (— cos x) (— cos y) — (— sin x) sin y 

= cos x cos y + sin x sin y. 

Note 3. — The four fundamental formulae just proved are very important, and 
must be committed to memory. It will be convenient to refer to them as the ' x, y ' 
formulas. From any one of them, all the others can be deduced in the following 
manner : 

Thus, from cos (x — y) to deduce sin (x + y). We have 

cos (x — y) = cos# cosy + sin# siny . . . . (1) 

Substitute 90°— x for # in (1), and it becomes 

cos {90°— (x + y)\ = cos (90°— x) cosy + sin (90°— x) sin y. 

. \ sin (x + y ) = sin x cos y + cos x sin y. (Art. 29) 

The student should make the substitutions indicated 
below, and satisfy himself that the corresponding results 
follow : 



54 PLANE TBIGONOMETRY. 

From 

sin (x + y) to deduce cos (x + 2/) substitute (90°+#) for a?. 

" " " cos(a-y) " (90°- a?) for a. 

cos (a? + y) " " sin (a + 2/) " (90° + x) for a. 

" " " sin (a - y) « (90°- a;) for x. 

" " " cos (a? — 2/) " —yiovy. 

etc. etc. etc. 

EXAMPLES. 

1. To find the value of sin 15°. 

sinl5°=sin(45 o -30°) 

= sin 45° cos 30° - cos 45° sin 30° 

1 V3_J_ 1 

V2' ^ ' V2*2 

V3-1 









2V2 





Show that 


sin 75° = 


V3 + 1 

2V2 


3. 


Show that 


cos 75°- 


V3-1 

2V2 


4. 


Show that 


cosl5° = 


V3 + 1 



2V2 

3 5 

5. If sin x = ~j and cos y = — find sin (# + ?/) and 

<*»(*-*)• ^ns. §?, and g. 

65 65 

6. If sin x = -, and cos 2/ = -, find sin (a; + y) and 
C ° S(a; - 2/) - ,1ns. 1, and # 



FORMULA FOR TRANSFORMATION. 55 

45. Formulae for the Transformation of Sums into Prod- 
ucts. — From the four fundamental formulae of Arts. 43 
and 44 we have, by addition and subtraction, the following: 

sin (x + y) + sin (x — y) = 2sinxcosy ... (1) 

sin (x -j- y) — sin (x — y) = 2 cos a; siny . . . (2) 

cos(x + y) + cos (x — y) = 2cos x cosy . . . (3) 

cos (x — y) — cos (x -f?/) = 2 sin x siny . . . (4) 

These formulae are useful in proving identities by trans- 
forming products into terms of first degree. They enable 
us, when read from right to left, to replace the product of a 
sine or a cosine into a sine or a cosine by half the sum or 
half the difference of two such ratios. 

Let x + y = A, and x — y = B. 

.-. x = ±(A + B), and y = i(A-B). 

Substituting these values in the above formulae, and 
putting, for the sake of uniformity of notation, x, y instead 
of A, B, we get 

sin# + sin?/ = 2sin-J- (x + y) cos-J-(£ — y) . . (5) 

sin x — sin y = 2 cos % (x + y) sin -J- (x — y) . . (6 ) 

cos x + cos y = 2 cos i (x + y) cos \ {x — y) . . (7) 

cos y — cos x = 2 sin -J (# + #) sin ^ (# — ?/) . . (8) 

The formulae are of great importance in mathematical 
investigations (especially in computations by logarithms) ; 
they enable us to express the sum or the difference of two 
sines or two cosines in the form of a product. The student 
is recommended to become familiar with them, and to com 
mit the following enunciations to memory : 

Of any two angles, the 

Sum of the sines = 2 sin £ sum • cos-Jdiff. 
Diff. " " " =2cos£sum.sin£diff. 



56 PLANE TRIGONOMETRY. 

Sum of the cosines = 2 cos % sum • cos £ diff. 
Diff. " " " =2sinisum.sin^diff. 

EXAMPLES. 

1. sin5cccos3ce = |-(sin8cc + sin2#). 

For, sin5ajcos3cc = ^-{sin (5cc + 3a;) + sin (5cc — See) } 
== ^ (sin 8 x + sin 2 x) . 

2. Prove sin6 sin 3 (9 =^ (cos2<9 - cos40). 

3. " 2sin0cos<£ = sin (6 + <£) + sin (0- <£). 

4. " 2sin20cos3<£ = sin(20 + 3<£) + sin (2<9-3<£). 

5. " sin 60° + sin 30° = 2 sin 45° cos 15°. 

6. " sin 40° -sin 10° =2 cos 25° sin 15°. 

7. " sinlO0+sin60=2sin8(9cos20. 

8. " sin8a— sin4a = 2cos6asin2a. 

9. " sin3cc + since = 2 sin 2 x cos ce. 

10. " sin 3 x — since = 2 cos 2 ce sin #. 

11. " sin4cc + sin2cc = 2sin3cecosct\ 

46. Useful Formulae. — The following formulae, which 
are of frequent use, may be deduced by taking the quotient 
of each pair of the formulae (5) to (8) of Art. 45 as follows : 

-. since + sin?/ __ 2sin|-(ce + y) cos^(x-y) 
since— sin y 2cos|-(cc + y) sin -J-(cc — y) 

= tan|(a; + 3/) cot^(x-y) 

= tanj(g + y) . (Art 24) 

tan £ (x — y) 

The following may be proved by the student in a similar 

manner : 

o sin x + sin y , , , , x 

2. 2- ^ = tan£(z + ?/), 

coscc + cosy 



THE TANGENT OF TWO ANGLES. 57 

o sin x 4- sin y >+, v 

3. -£ ^ = cot|(a?-y), 

cosy — cos a; 

, sin sc — sin?/ . . , , N 

cos x + cos 2/ 

5. sina ? - S in^ coH(a; + y)} 

cos 2/ — cos a; 

6. 0Mg + eM y = coti(s + y)cotKs-y). 

cos?/ — cos a; 

47. The Tangent of the Sum and Difference of Two Angles. 

— Expressions for the value of tan (a? + y), tan (& — y), etc., 
may be established geometrically. It is simpler, however, 
to deduce them from the formulae already established, as 
follows : 

Dividing the first of the 'x, y ? formulae by the second, 
we have, by Art. 23, 

. , \ _ s ^ n ( x + V) __ sin x CQS V + cos x s ^ n 2/ 

cos (# + ?/) cos # cos ?/ — sin x sin ?/ 

Dividing both terms of the fraction by cos a? cosy, 

sin x cos y cos x sin ?/ 

, x cos x cosv cos # cos v 

tan (a; + y) = - : r-2 

v ' cos x cosy sina;sm?/ 

cos x cos ?/ cos x cos ?/ 

tans + tany (Art 23) . . . (1) 
1 — tan x tan y 

In the same manner may be derived 

to(»-y)- 1 t r?--» n * (2) 

1 + tana; tan?/ 

Also, cotrx + y)^ 00 ^ 00 ^" 1 (3) 

cot# + cot?/ 

and cotO-y) = cota;cot y + 1 (4) 

cot?/ —cot a; 



58 



>LANE TRIGONOMETRY. 



EXERCISES. 



Prove the following : 



1. 


tan (x + 45°)- ™ "^ . 
1 — tan x 


2. 


tan(x 45°)- tan *"~ 1 . 
v 1 + tana 


3. 


sin (x + ?/) tan x + tan?/ 
sin (x — y) tan # — tan 2/ 


4. 


cos (x — y) _ tan a? tan y + 1 



cos (# + ?/) 1 — tan x tan ?/ 

5. sin (x + y) sin (x — y)= sin 2 x — sin 2 ?/ 

= cos 2 ?/ — cos 2 #. 

6. cos (x + y) cos (x — y) = cos 2 x — sin 2 ?/ 



= cos 2 ?/ — sin 2 #. 



7. 

8. 
9. 



tana? ± tan?/ = * — ±Jll. 

cos x cosy 

, , , sin (y ± x) 

sin a; sin?/ 



sin2# cos2# 



= sec#. 



sm# 



cosx 



10. If tan x = ^ and tan y = i, prove that tan (x + 2/) = f ? 
and tan (x — y) = f- . 

11. Prove that tan 15° = 2 - V& 

12. If tan#=f, and tan?/ = T 1 T > prove that tan (a; + ?/) =1. 
What is (x + ?/) in this case ? 

48. Formulae for the Sum of Three or More Angles. — Let 

x, y, z be any three angles ; we have by Art. 43, 



EXAMPLES. 59 

sin (x + y + 2) = sin (x + 2/) cos 2 + cos (x + y) sin 2 
= sin a; cos y cos 2 + cos x sin 2/ cos z 

» + cos # cos y sin 2 — sin # sin y sin 2 . . (1) 

In like manner, 
cos (x + y + 2) = cos # cos 2/ cos 2 — sin x sin 2/ cos 2 

— sin x cos 2/ sin z — cos aj sin 2/ sin z . . (2) 

Dividing (1) by (2), and reducing by dividing both terms 
of the fraction by cos x cos y cos 2, we get 

, , . N tan cc + tan ?/ + tan 2 — tan a? tan?/ tan 2 

tm(x+y+z) = - ■ ^- J 2 (d) 

1 — tana; tan?/ — tan?/ tanz — tanz tana; 

EXAMPLES. 

1. Prove that sin x + sin y + sin z — sin (x + y + z) 

= 4sin|(a; + 2/) sin£(y + 2) sin J(« + x). 
By (6) of Art. 44 we have 

sina; — sin (x + y + z) = —2cos%(2x + y + z) sm%(y + z) 9 
and sin2/ + sin2 = 2sin|-(2/ + 2) cos £(?/ — «)• 

.\ sina; + sin y + sin2 — sin (a; + y + 2) 

= 2sin£(2/+2) cos £(2/— 2) -2cosi(2x+y+z) sin J (3/ +2) 
=2sini(t/ + 2) {cos£(y - 2) — cos£(2a; + 2/ + 2) } 
= 2sin|(2/ + 2) 2sin-J-(a3 + ?/) sin -J- (a; + 2) 
==4sin£(a; + 2/) sinf (y + 2) sin|(2 + x). 
Prove the following : 

2. cos x + cos y + cos 2 + cos (x + y + z) 

= 4cos + (?/ + 2) cos -J- (2 + a;) cos | (a; + 2/). 



60 PLANE TRIGONOMETRY. 

3. sin (x + y — z) = sin a; cos y cos z + cos a; siny cos z 

— cos x cosy sin z + sin a? siny sinz. 

4. sina; + sin?/ — sinz — sin (x + y — z) 

= 4sin£(a; — z) sin£(y — z) sin£(a; + y). 

5. sin (y — z) + sin (z — a?) + sin (x — y) 

+ 4sin|-(2/— z) sin^(z — x) sin J (a? — 2/) = 0. 

49. Functions of Double Angles. — To express the trigo- 
nometric functions of the angle 2 a; in terms of those of the 
angle x. 

Put y = x in (1) of Art. 42, and it becomes 
sin 2x= sin x cos a? + cos x sin a?, 

or sin2# = 2sina;cosa: (1) 

Put y = x in (2) of Art. 42, and it becomes 

cos2# = cos 2 a; — sin 2 a; (2) 

= l-2sin 2 aj (3; 

or = 2cos 2 # — 1 (4) 

Put y = x in (1) and (3) of Art. 47, and they become 



2 tan a; 
1 — tan 2 x 



**2" = ^t£z ( 5 > 






oot2s = cot2 *- 1 (6) 

2cota; v ' 

Transposing 1 in (4), and dividing it into (1), we have 

sin 2 a; , ,rr\ 

— - = tan x (7) 

l + cos2* 






EXAMPLES. 61 

Note. — These seven formulae are very important. The student must notice that 
x is any angle, and therefore these formulae will be true whatever we put for x. 

Thus, if we write - for x, we get 

sin x = 2 sin - cos - (8) 

2 2 

cos x = cos 2 -- sin 2 f (9) 

or = l-2sin 2 - = 2cos 2 --l (10) 

2 2 

and so on. 

EXAMPLES. 

Prove the following : 
1. 2cosec2# = sec#cosec#. 






o cosec 2 # 

2. = sec 2 x. 

cosec J #— 2 

. — =sin2a;. 

1 + tan 2 a; 

A 1— tan 2 a; 

4. — = cos 2 x. 

l + tan J £ 

5. tan x + cot x = 2 cosec 2 x. 

6. cot x — tan x = 2 cot 2 x. 

„ sin# , x 

7. =tan-- 

1 + cos a; 2 

Q sin# ±x 

o. =cot — 

1 — cos# 2 

9. Given sin 45°= A:; find tan 22*°. Ans. V2-1. 

V2 

3 24 24 

10. Given tansc = -; find tan2#, and sin2#. — , — 

4' 7 ? 25 

50. To Express the Functions of 3 a; in Terms of the Func- 
tions of x. 

Put y = 2x in (1) of Art. 42, and it becomes 



62 PLANE TRIGONOMETRY. 

sin3#= sin (2x + x) 

= sin2a;cosa; + cos2a;sina; 

= 2 sin # cos 2 a; + (1 — 2 sin 2 x) sin # (Art. 49) 

= 2 sin a; (1 — sin 2 ^) + sin a; — 2sin 3 a> 

= 3sin# — 4sin 3 a?. 
cos 3 a? = cos (2x + x) 

= cos 2 x cos x — sin 2 x sin # 

= (2cos 2 x — 1) cos a; — 2 sin 2 a; cos a; (Art. 49) 

= 4 cos 3 x — 3 cos x. 

tan3#= tan2a? + x 

tan 2x + tan a; 
1 — tan2x tana; 

2 tan a; 



tan 2 a; 



+ tana; 



-j 2 tan 2 a? 



1 — tan 2 a; 

_ 3 tan x — tan 3 x 
1 — 3tan 2 a? 

EXAMPLES. 

Prove the following : 

Ism o X r* o i -I 

. = 2cos2a; + l. 

sinx 

2. sin3 ^- siM =tanx. 
cos 3 a; + cos a; 

3. 8in3a? + cos3g = 2sin2 !B +l. 

cos a; — sin a; 



FUNCTIONS OF THE HALF ANGLE. 63 



4. H = cot2a% 

tan 3 x — tan x cot x — cot 3 x 

5. 1 - cos3x = (l+2cosx)\ 

1 — cos a; 



51. Functions of Half an Angle. — To express the func- 

x 

2 



tions of - in terms of the functions of x. 



Since cos x = 1 — 2 sin 2 -, 

or = 2cos 2 --l . . . [Art. 49, (10)] 

Zi 

• nX 1 — COS.T /i > 

,\ sm 2 - = (1; 

2 2 w 

and cos 2 - = 1 + COSa? (2) 

2 2 v } 



x /I — cos a; 

l 2 = ± \Hr 



= ±\ g ( * 



-, a; /l + cos a; /ylv 

and cos- = ±^-^ (4) 

t> t. • . . 03 /l — cosaj , 1— cosx /cr . 

By division, tan - = ± x - = ± : . . (5) 

2 \l + cos# smx 

By formulae (3), (4), and (5) the functions of half an 
angle may be found when the cosine of the whole angle is 
given. 

52. If the Cosine of an Angle be given, the Sine and the 
Cosine of its Half are each Two-Valued. 

By Art. 51, eacli value of cos a; (nothing else being known 

about the angle x) gives two values each for sin- and cos-, 
one positive and one negative. But if the value of x be 



64 



PLANE TRIGONOMETRY. 



given, we know the quadrant in which - lies, and hence 

we know which sign is to be taken. 

x 

Thus, if x lies between 0° and 360°, - lies between 0° and 

9 2 

x 
180°, and therefore sin- is positive; but if x lies between 

360° and 720°, x - lies between 180° and 360°, and hence 

2 
sin- is negative. Also, if x lie between 0° and 180°, cos- 

— A 

is positive; but if x lie between 180° and 360°, cos- is 

negative. 

The case may be investigated geometrically thus : 

Let OM = the given cosine (radius being unity, Art. 16) 

= cos x. Through M draw PQ per- 
pendicular to O A ; and draw OP, OQ. 

Then all angles whose cosines are 

equal to cosx are terminated either 

by OP or OQ, and the halves of these 

angles are terminated by the dotted 

lines Op, Oq, Or, or Os. The sines 

of angles ending at Op and Oq are 

the same, and equal numerically to 

those of angles ending at Or and Os ; but in the former case 

they are positive, and in the latter, negative ; hence we 

obtain two, and only two, values of sin- from a given value 
of cos x. 

Also, the cosines of angles ending at Op and Os are the 
same, and have the positive sign. They are equal numeri- 
cally to the cosines of the angles ending at Oq and O?', but 
the latter are negative ; hence we obtain two, and only two, 

values of cos- from a given value of cos#. 

Also, the tangent of half the angle whose cosine is given 
is two-valued. This follows immediately from (5) of Art. 
51. 




FUNCTIONS OF THE HALF ANGLE. 



65 



53. If the Sine of an Angle be given, the Sine and the 
Cosine of its Half are each Four- Valued. 



We have 



and 



By addition, 






and 



and 



X X 

2sin-cos- = sin a; . . . 

2 2 


(Art. 49) 


• 9 X . X -4 

SUT- + COS--= 1 .... 

2 2 


(Art. 23) 


sin j + cos f ) =14- sin x. 




x x\^ 
sin cos - ] =l-sina;, 

2 2) 




. sin - + cos - = ± Vl + sin x 
2 2 


. . (1) 


sin- — cos-= ± Vl — sina; 
2 2 


. . (2) 


= ± Vl + sina; ± Vl — sina; 


. . (3) 



2sm- 



2cos-= ± Vl + sina;T Vl 



sma; 



(4) 



Thus, if we are given the value of since (nothing else 

being known about the angle x), it follows from (3) and 

x x 

(4) that sin- and cos- have each four values equal, two 

by two, in absolute value, but of contrary signs. 
The case may be investigated geometrically thus : 
Let OX = the given sine (radius being unity) = sina;. 

Through N draw PQ parallel to OA; 

and draw OP, OQ. Then all angles 

whose sines are equal to sina; are 

terminated either by OP or OQ, and 

the halves of these angles are termi- 
nated by the dotted lines Op, Oq, Or, 

or Os. The sines of angles ending 

at Op, Oq, Or, and Os are all different 




66 PLANE TBIGONOMETBY. 

in value ; and so are their cosines. Hence we obtain four 

X X 

values for sin-, and four also for cos-, in terms of x. 

2' 2 

When the angle x is given, there is no ambiguity in the 

x 

calculations ; for* - is then known, and therefore the signs 

Z 

X X 

and relative magnitudes of sin- and cos r are known. Then 

Z Z 

equations (1) and (2), which should always be used, im- 
mediately determine the signs to be taken in equations (3) 
and (4). 

Thus, when - lies between — 45° and + 45°, cos- > sin-, 
' 2 2 2 

and is positive. 

Therefore, (1) is positive, and (2) is negative ■ and hence 
(3) and (4) become 

2 sin- = Vl + sin# — Vl — sin x, 

Z 



2 cos - = Vl + sinx + Vl — sin x. 

Z 

When ^ lies between 45° and 135°, sin- > cos-, and is 

2 2 2 

positive. 

Therefore (1) and (2) are both positive; and hence (3) 
and (4) become 

2 sin- = Vl + sin a; + Vl — sin#, 

z 



2cos-= Vl + sin a;— Vl — sin x. 

z 

And so on. 

54. If the Tangent of an Angle be given, the Tangent 
of its Half is Two- Valued. 

2tan i 
We have tan = \ (Art. 49) 

1-tan 2 " 

2 



FUNCTIONS OF THE HALF ANGLE. 67 



Put tan- = x: thus 

2 ' 



(l-& 2 )tan0 = 2a?, 
._*. 2 



tan0 



:»=1. 



2 tan<9 

Q 

Thus, given tan0, we find two unequal values for tan-, 

one positive and one negative. 

This result may be proved geometrically) an exercise which 
we leave for the student. 

55. If the Sine of an Angle be given, the Sine of One- 
Third of the Angle is Three-Valued. 

We have sin 3 x = 3 sin x — 4 sin 3 x . . (Art. 50) 

Q 

Put • x = -, and we get 

3 

8 6 

sin = 3 sin 4 sin 3 -, 

3 3 

a cubic equation, which therefore has three roots. 



EXAMPLES. 

1. Determine the limits between which A must lie to 
satisfy the equation 

2sinA=s— Vl + sin2A— Vl — sin 2 A. 

By (1) and (2) of Art. 53, 2 sin A can have this value . 
only when 

sin A + cos A = ■— Vl + sin 2 A, 



and sin A — cos A = — Vl — sin2A; 

i.e., when sin A > cos A and negative. 



68 PLANE TRIGONOMETRY. 

Therefore A lies between 225° and 315°, or between the 
angles formed by adding or subtracting any multiple of 
four right angles to each of these ; i.e., A lies between 

2mr+— and 2mr + — , 
4 4 

where n is zero or any positive or negative integer. 

2. Determine the limits between which A must lie to 
satisfy the equation 



2cosA = Vl + sin2A — Vl — sin2A. 

By (1) and (2) of Art. 53, 2 cos A can have this value 
only when 

cos A + sin A= Vl + sin2A, 



and cos A — sin A = — Vl — sin 2 A ; 

i.e., when sinA>cosA and positive. 
Therefore A lies between 

2 n-K 4- - and 2 mr -\ — -, 
4 4 

where n is any positive or negative integer. 

3. State the signs of (sin + cos 6) and (sin — cos 6) 
when 6 has the following values: (1) 22°; (2) 191°; 
(3) 290°; (4) 345°; (5) -22°; (6) -275°; (7) -470°; 
(8) 1000°. 

Ana. (1) +, -; (2) -, +; (3) -, -; (4) +, -j 
(5) +, -; (6) +, +; (7) -, - j (8) -,-.. 

4. Prove that the formulae which give the values of 

sin- and of cos- in terms of sin a; are unaltered when x 

2 2 

has the values 



(1) 92°, 268°, 900°, 4ft7r + f7r, or (4n + 2)7r-f7r; 

(2) 88°, - 88°, 770°, - 770°, or £mr ± 



71 

8 



i 



VALUES OF SPECIAL ANGLES. 69 

5. Find the limits between which A must lie when 



2sinA = Vl + sin2A — Vl — sin2A 

1 the Values of the Functions of 

(4), and (5) of Art. 51, put x = 45°. Then 



56. Find the Values of the Functions of 22|°. — In (3), 



sin 22^ 



o /l — CC 



cos 45° 



V2 



2i° = / l + cos45° _ V2 4- V2 



■v 



tan 22*°= 



2 
• cos 45° 



sin 45° 



= V2 - 1. 



Since 22^-° is an acute angle, its functions are all positive. 

The above results are also the cosine, sine, and cotangent 
respectively of 67^-°, since the latter is the complement of 
22^° (Art. 15). 

57. Find the Sine and Cosine of 18°. 

Let a =18°; then 2^ = 36°, and 3 x = 54°. 

.-. 2x + 33 = 90°. 

.\ sin 2 a; = cos3# (Art. 15) 

.-. 2 sin x cos x = 4 cos 3 x — 3 cos x . . (Art. 50) 

or 2sin# = 4cos 2 # — 3 

= 1 — 4 sin 2 x. 

Solving the quadratic, and taking the upper sign, since 
sin 18° must be positive, we get 

V5-1 



sin 18° = 



Also, cos 18°= Vl-sin 2 18° = 



VlO + 2^/5 



Hence we have also the sine and cosine of 72° (Art. 15). 



70 PLANE TBIGONOMETBY. 

58. Find the Sine and Cosine of 36°. 

cos 36° - 1 - 2 sin 2 18° . . . [ (3) of Art. 49] 

= 1 6-2V5 = V5 + 1 
8 4 

o All0-2-\/5 



.-. sin 36°= Vl-cos 2 36' 

The above results are also the sine and cosine, respec- 
tively, of 54° (Art. 15). 

Otherwise thus: Let x = 36°; then 2a=72°, and 3^=108°. 

.-. 2^ + 3^ = 180°. 

.\ sin2a; = sin3a (Art. 29) 

2 sin x cos x = 3sin# —4 sin 3 a;, 
2 cos a; = 3 — 4sin 2 # 
= 4cos 2 # — 1. 

a i • ± V5 + 1 
Solving, cos x = — 

4 

But 36° is an acute angle, and therefore its cosine is 
positive. 

o/?o V5 + 1 
.-. cos 36 = - — - — 
4 

59. If A + B + C = 180°, or if A, B, C are the Angles 
of a Triangle, prove the Following Identities: 

(1) sin A + sinB + sin C =4cos— cos — cos — 

Ld Li Ld 

ABC 

(2) cosA + cosB + cos C = 1 + 4sin — sin— sin-- 

Z A Z 

(3) tan A + tanB + tanC = tan A tan B tanC. 






ANGLES OF A TBIANGLE. 71 



We have A + B + C = 180°. 
.-. sin(A+B)=sinC, and sin — ^ — = cos— • (Arts. 15 and 30) 

Now sinA+ sinB =2sin — ~ — cos — ~ — . (Art. 45) 

= 2cos^cos^-=-? . . (Art. 15) 

- - 

c c 

and sinC = 2sin-cos- . . . . (Art. 49) 

— z 

= 2cos^-±-?cos^ . . (Art. 15) 

.'. sin A + sinB + sinC = 2cos — cos — — — f-2cos-cos — t— 

2 2 2 2 

o C/ A-B f A + B 

= 2cos- cos 1- cos 



2^2 '2 

== 2cos— ( 2cos— cos—]. (Art. 45) 

A A B C M v 

= 4 cos— cos — cos — .... (1) 

2 2 2 w 

Again, cos A + cos B = 2 cos — i — cos — ~ — . (Art. 45) 






= 2sin-cos — 

2 2 ' 

and cosC = l-2sin 2 ^ .... (Art. 49) 

C / A — B C\ 

.-. cosA + cosB + cosC= l + 2sin- (cos — sin- ] 



G( A-B „A+B\ 

2 J 



= 1 + 2 sin ^ /"cos ^— — - cos 



= l + 4sin-sin5 s in§ . . (2) 
2 2 2 w 



72 



PLANE TRIGONOMETRY. 



Again, tan (A + B) = — tan C (Art. 30) 

tanA + tanB ,» . ,_. 
"1 — tan A tan B' ^ *' ' 

.-. tanA + tanB = — tan C (1 — tan A tanB). 

.\ tanA + tanB + tan C = tan A tanB tan C .... (3) 

Note. — The student will observe that (1), (2), and (3) follow directly from 
Examples 1 and 2, and formula (3), respectively, of Art. 48, by putting 

A + B + C = 180°. 



EXAMPLES. 

Prove the following statements if A + B + C = 180° : 

1. cos(A + B-C) = -cos2C. 

A B C 

2. sin A + sinB — sinC = 4sin— sin— cos-« 

■ 2 2 2 

3. sin2A + sin2B + sin2C = 4sinA sinB sinC. 

4. sin2A + sin2B — sin2C = 4sinC cosAcosB. 

5. tan7 A — tan4 A — tan3 A = tan7 A tan4 A tan3 A. 

ABC 

6. sinA — sinB + sinC = 4sin— cos- sin — 

2 2 2 

7. cot — h cot — h cot - = cot — cot — cot — 

222 222 

8. tan A — cotB = secAcosecB cosC. 

60. Inverse Trigonometric Functions. — The equation 
sin == x means that 6 is the angle whose sine is x ; this 
may be written 0= sin -1 ^ where sin -1 # is an abbreviation 
for the angle (or arc) whose sine is x. 

So the symbols cos -1 ^ tan -1 #, and sec -1 !/, are read "the 
angle (or arc) whose cosine is x" "the angle (or arc) whose, 
tangent is xf and "the angle (or arc) whose secant is y" 
These angles are spoken of as being the inverse sine of x, the 



INVERSE TRIGONOMETRIC FUNCTIONS. 73 

inverse cosine of x, the inverse tangent of x, and the inverse 
secant ofy, respectively. Such expressions are called inverse 
trigon ometric functions. 

Note. — The student must be careful to notice that —1 is not an exponent, sin" 1 a? 

is not (sin re) -1 , which = 

sin a; 

Notice also that sin" 1 — = cos -1 - is not an identity, but is true only for the par- 
ticular angle 60°. 

This notation is only analogous to the use of exponents in multiplication, where 
we have a~ l a = a° = 1. Thus, cos -1 (cos x) — x, and sin (sin -1 x) = x; that is, cos" 1 
is inverse to cos, and applied to it annuls it; and so for other functions. 

The French method of writing inverse functions is 
arc sinx, arc cosx, arc tanx, and so on. 



EXAMPLES. 

1. Show that 30° is one value of sin"" 1 ^-. 

We know that sin 30°= -J-. .-. 30° is an angle whose*sine 
is^; or 30°= sin- 1 i 

2. Prove that tan" 1 ! + tan" 1 ! = 45°. 

tan" 1 } is one of the angles whose tangent is f, and tan -1 -!- 
is one of the angles whose tangent is i. 

Let a = tan- 1 ^- ? and /3 = tan- 1 ^; 

then tana = |- and tan/3 = ^. 

Now tan(« + /?) = > n " + tan ff . . . (Art. 47) 
1 — tan a tan/3 



= -i±i- = l 

But tan 45°= 1, .-. « + = 45°; 
that is, tan" 1 f + tan- 1 J = 45°. 



74 PLANE TRIGONOMETRY. 

3. Prove that tan" 1 ^ + tan" 1 ?/ = tan" 1 — iiL 

1 — xy 

Let tan" 1 x = A. . \ tan A = x. 
tan" 1 y = B. .\ tanB = 2/. 

Now tan (A + B) = tanA + tanB 

v } 1-tanAtanB 

_x + y 



1 — xy 



.-. A + B = tan- 1 -^±^-. 
1 — xy 

,\ tan" 1 a; + tan" 1 ?/ = tan" 1 ^ ~*~ ^ - 

Aiyr relations which have been established among the 
trigonometric functions may be expressed by means of the 
inverse notation. Thus, we know that 



4. cos x = Vl — sin 2 #. 



This may be written x = cos -1 Vl — sin 2 a; . . (1) 

Put sin x = 6 ; then x = sin" 1 0. 



Thus (1) becomes sin" 1 6 = cos -1 Vl — 2 . 

5. By Art. 49, cos20 = 2cos 2 0- 1, 
which may be written 26 = cos" 1 (2 cos 2 6 — 1). 

Put cos0 = cc. .-. 2cos~ 1 # = cos" 1 (2^ — 1). 

6. By Art. 49, sin 2 6 = 2 sin 6 cos 0, 
which may be written 20 = sin -1 (2 sin 6 cos 6) . 



Put sin 6 = x. .-. 2sin" 1 #= sin" 1 (2 a; Vl — x 2 ). 



TABLE OF USEFUL FORMULAE. 



75 



7. Prove 
8. 



sin" 1 a? = cos"" 1 Vl — x 2 = tan" 1 ■ 



vr 



tan" 1 a; = sin -1 



. — ons -1 



vr 



COS" 



Vl + ar" 



9. 

10. 
11. 

12. 

13. 
14. 



" 2tan~ 1 a; = tan- , -^~ 

1 — or 

■ 

" sin (2 sin" 1 a) = 2 a; Vl - a? 2 . 

tan- 1 - + tan- 1 - = 2. 
7 6 4 

« cos- 1 i + 2sin- 1 l = 120°. 

2 2 

« cot- 1 3 + cosec- 1 V5 = -- 

4 

" 3 sin" 1 a; = sin" 1 (3 x — 4 or 3 ) 



61. Table of Useful Formulae. — The following is a list 
of important formulae proved in this chapter, and summed 
up for the convenience of the student : 

1. sin (x + y) = sin x cos y + cos x sin y . . (Art. 43) 

2. cos (a; + y) = cos a; cos y — sin a; sin y. 

3. sin (x — y) = sin a? cos ?/ — cos x sin 2/ . . (Art. 44) 

4. cos (x — y) = cos a; cos y + since sin y. 

5. 2 sin x cos 2/ = sin (x + y) + sin (x — y) . (Art. 45) 

6. 2 cos a; sin y =sin (x + y) — sin(x — y). 

7. 2cos#cos2/ =cos (x + y) + cos (a; — y). 

8. 2sina;sin2/ = cos (x — y) — cos (x + y) . 

9. since + sin?/ = 2sin|-(a; + y) cos|-(a; — y). 
10. since— siny = 2cos|(# + y) sin£(a; — y). 



PLANE TRIGONOMETRY, 



11. cos# + cos y — 2cos-^(# + ?/) cos|-(a; — y). 

12. cosy — co$x = 2 sin -J- (as + ?/) sin|-(cc —y). 

13 sin a; + sin?/ = fa»^(a? + y) ^^ 4g v 

sin a; — sin 2/ tan J (a; — 2/) 

14. tan + ?/) = tan ^ + tan ^ (Art. 47) 

1 — tana; tan?/ 



15. tan (x — y) 

16. cot (x + y) 



tan x — tan y 
1 + tanxtan?/ 

cot x cot ,y — 1 
cot# + cot?/ 



-. ~ , / x cot x cot v + 1 

17. cot (# — y)= ^— - — 

cot a; — cot?/ 



18. tan (a? ±45°) = 



tana; T 1 



(Art. 47) 



tan x ±1 

19. sin (x-\- y) sin (x — y) = sin 2 a? — sin 2 ?/ = cos 2 ?/ — cos 2 #. 

20. cos (a?+ y) cos (# — ?/) = cos 2 a; — sin 2 ?/ = cos 2 ?/ — sin 2 a;. 

21. tan a; ±tany = sin ( a;± ^. 

COS X COS ?/ 

22. cotx±coty= sin ^ ±x ^ 

sin a; sin?/ 



23. sin 2 # = 2 sin x cos a? = 



2 tan a; 



1 + tan 2 # 

24. cos 2 a; = cos 2 a; — sin 2 # = 1 — 2sin 2 # = 2cos 2 a; — 1 

_ 1 — tan 2 a? 
1 + tan 2 a? 

ok 1 — cos 2 a; 2 sin 2 a; 2 

25. — - = — = tan 2 a\ 

1 + cos 2 a; 2cos J a; 



(Art. 49) 







EXAMPLES. 


26. 


tan 2 a; 


2 tan a; 
1 — tan 2 a; 


27. 


cot 2 a; 


cot 2 a; — 1 
2 cot a; 


28. 


sin 3 x 


= 3 sin x — 4 sin 3 x 



77 



(Art. 50) 

29. cos 3 x = 4 cos 3 x — 3 cos #. 
3 tan x — tan 3 # 



30. tan 3 a; : 



1 — 3 tan 2 x 



31. sin 2 - = 1 ~^ osa; (Art 51) 

32. cos 2 * = * + <***. 

2 2 

33. tan~ 1 a; + tan- 1 2/ = tan- 1 ^+^- .... (Art. 60) 



EXAMPLES. 

1 . 2 

1. If sin a = -, and sin /3 = -, find a value for sin ( a+ /?) 
o o 

andsin(a-/3). ^ V5 + 4V2 . V5 - 4 V : 



9 ' 9 

2. If cos a = -, and cos /3 = — , find a value for sin ( a + j8) 

o 41 

and cos (<* + /?). ^ 156 133 

205' 205 

3 2 

3. If cos a = -, and cos /} = -, find a value for sin (a + /?) 

4 5 

and sin (a -0). , 2 V7 + 3 V 21 2 V7 - 3 V 21 



Ans, 



20 20 



4 3 

4. If sin a — -, and sin ft =* -, find a value for sin (« + /?) 

5 5 

and cos (a + P). A 1 24 



78 PLANE TRIGONOMETRY. 

5 

5. If since =.6, and sin /? = — , find a value for sin (a— /?), 

and cos (a + /?). +n qo 

.4ns. — , — • 
65 65 

6. If since= — ±, and sin/? = — = , show that one value 

V5 Vio 

of a + /? is 45°. 

7. Prove cos + cos 3 = 2 cos 20 cos 0. 

8. " 2 cos a cos ft = cos (a — /3) + cos (a + /3). 

9. " 2sin30cos50 = sin80 - sin20. 

10. " 2cosf0cos- == cos (9 + cos 20. 

11. " sin 40 sin = £(cos30 - cos 50). 

12. " 2 cos 10° sin 50° = sin 60° + sin 40°. 

13. Simplify 2 cos 2 cos - 2 sin 4 sin 0. 

-4ns. 2 cos 30 cos 20. 

14. Simplify sin — cos sin — cos — — cos 4 sin 2 0. 

r J 2 2 2 2 

Prove the following statements : 

15. cos 3 a — cos 7 a = 2 sin 5 a sin 2 #. 

16. sin 60° + sin 20° = 2 sin 40° cos 20°. 

17. sin30 + sin50 =2sin40cos0. 

18. sin70~-sin50 = 2 cos 6 sin 0. 

19. cos50 + cos90 = 2cos70cos20. 

20. sin20 + sin0 =tan^. 
cos0 + eos20 2 

21. cos (60° + A) + cos (60° - A) = cos A. 






EXAMPLES. 79 

22. cos (45° + A) + cos (45° - A) = V2 cos A. 

23. sin (45°+ A) -sin (45°- A) = V2sinA. 

24. cos20 + cos40 = 2cos30cos0. 

25. cos40-cos60 = 2sin50sin0. 

26. cos + cos 3 + cos 5 + cos 7 5 = 4 cos cos 2 cos 4 0. 

Or* I. , *. /3 COS (OC ~ /J) 

27. cot a + tan /? = — r— * ^ • 

since cos/5 

oo i. /> cos (a + B) 

28. cot a — tan /? = — ^ — L_ti/ - 

sm cc cos /? 

29. sin(A-45 ) = SlnA ~~ COsA . 

V2 . 

30. V2sin(A + 45°) = sinA + cosA. 

31. cos (A + 45°) + sin (A - 45°) = 0. 

32. tan (*"*) + tan * = tan ft 

1 — tan (0 — <£) tan <£ 

33. tan (0 + 0) + tan <fr = tan 
1 + tan (0 + 0) tan0 

34. cos (0 + 0) - sin (0 - 0) = 2sin f- - flVos f- - 0*V 

35. sin n6 cos + cos nO sin = sin (n + 1 ) 0. 

36. cot(e- 7 L] = eote + 1 . 

\ 4:) l-COt0 

37. tan (<? --) + cot (<? + -) = 0. 

38. cotfe--\ + ta.nfe + -\ = 0. 

39. tan(n + l)^-tann» =tan ^ 
1 4- tan (n + 1) <f> tan n<t> 



80 PLANE TRIGONOMETRY. 

\ 

40. If tana; = 1, and tan y = — -, prove that 

V3 

tan (x + y) = 2 + V3. 

41. If tana = , and tan/2 = , prove that 

m + 1 2m + 1 

tan(c6 + )8) = l. 

42. If tan a = m, and tan /? = n, prove that 

cos (a + /S) : 



V(l+m 2 )(l + n 2 ) 

43. If tan 6 = (a + 1), and tan <f> = (a — 1), prove that 

2cot(0 -</>) = a 2 . 

Prove the following statements : 

44. cos (x — y + z) = cos x cos y cosz + cosx sin?/sin2 

— sin x cos 2/ sin z + sin x sin y cos 2. 

45. sin (x — -y — 2) = sin x -f- sin t/ + sin z 

+ 4sin£(x- y) sin£(x— z) sin£(y + s). 

46. sin (x + ?/ — z) + sin (x + z — y) + sin (y + 2 — a;) 

= sin (x + 2/ + z) + 4 sin a; siny sin 2. 

47. sin2x + sin2?/ + sin2z — sin2 (x + y + z) 

= 4 sin (x + ?/) sin (y + 2) sin (2 + a:). 

48. cos 2 a; + cos 2 y + cos 22 + cos 2 (x + y + z) 

= 4 cos (x + y) cos (?/ + 2) cos (2 + x). 

49. cos (x + y — 2) 4- cos (y + z — x) + cos (2 + x — #) 

4- cos (x + 2/ + 2) = 4 cos x cos y cos 2. 



> 






EXAMPLES. 81 

50. sin 2 a: + sin 2 ?/ + sin 2 z + sin 2 (x + y + z) 

= 2J1 — cos (x + y) cos (y+z) cos {z+x) \. 

51. cos 2 a; + cos 2 ?/ + cos 2 z + cos 2 (x + y — z) 

= 2^1 + cos (x + y) cos (x—z) cos {y— z)\. 

52. cosa sin(?/ — z) + cosy sin(z— a) + cos2 sin (a?— y) = 0. 

53. sin a; sin (y — z) + siny sin (2 — x) + sing; sin (x — y) = 0. 

54. cos (x + y) cos (x — y) + sin (?/ + 2) sin (y — z) 

— cos (x + z) cos (a; — z) = 0. 

KK 2 — sec 2 nAC , /) 

55. = cos A v. 

secrO 

56. cos 2 0(l-r-tan 2 <9) = cos2ft 
cot 2 0-l 



57. cot 2 0: 



2 cot e 






58. Beo2tf = 52^±l. 

cot 2 0-l 

59. (sin I + cos | j = 1 + sin ft 

60. /"sin I - cos |Y = 1 - sin ft 



61. 1 + secg =2cos^. 
sec (9 2 

62 cos 20 _ 1 — tan 

l + sin20~~l + tan0' 

1 + tan- 
go cosfl _ 2 

" l-sintf -!_ tan f 






82 



64. 

65. 

66. 
67. 
68. 

69. 
70. 

71. 

72. 
73. 

74. 



PLANE TRIGONOMETRY. 

1 + sin# + cos# ,x 

—* — : l - = cot-- 

1 + sin# — cos& 2 



2 — sin2a? 

: 2 

2 + sin2a? 



cos 3 a? + sin 3 a; 
cosa; + sin a; 

cos 3 a; — sin 3 a; 
cos a; — sina? 2 

cos 4 6 — sin 4 6 = cos 2 0. 

sin30 cos30 



sin cos 
cos30 , sin30 



= 2. 



sin# 

sin4fl 
sin 20 

• 5-7T 

sm — 
12 



cosfl 
= 2cos20. 

5tt 



2cot20. 



cos 



12 



7T 7T 

Sm i2 C ° S i2 



= 2V3. 



tan (45°+ a) - tan (45°- x) = 2 tan 2 a?. 

tan (45°- x) + cot (45°- x) = 2 sec 2 a;. 

tan'(45°+*)-l = sin2a; , 
tan 2 (45°+a:) + l 

008(^ + 45°) = sec2a; _ tan2a; , 
cos (x — 45 ) 



75, 

76. tan# = 

77. tana== 
78. 



sina;+ sin2a: 
1 + eos# + cos2a; 

sin2# — sin a? 
1 — cosa? + cos 2 a; 



cos 3 a; 



cos a; 



= 2cos2aj — 1. 



EXAMPLES. 



83 






79. 3sina! - sin3a; Stan's. 

cos 3 x + 3 cos x 

cot 3 a; — 3 cot a; 



80. cot 3 x 



3cot 2 a-l 



81. 1 ~~ COs3a? =(l + 2cos3;) 2 . 
1 — cos a; 



82. sinx + cosx =: tan 2 a? + sec 2 a. 

cos a? — sina; 



83. 



cos 



2a; + cos 12a; cos7a; — cos3a ? 2sin4# _n 



cos 6 x + cos 8 x cos x — cos 3 x sin 2 a; 

84. sin2# sm2y = sin 2 (# + y) — sin 2 (# — y). 

85. tan 50° + cot 50° = 2 sec 10°. 

86. sin 3 x = 4 sin x sin (60° + x) sin (60° - a) . 



87. cot? -tan- = 2. 



8 



88. tan4fl= 4tang ( 1 -~ tan2g >. 

l-6tan 2 + tan 4 6> 

89. 2cos|= V2 + V2. 

o 

90. (3sin<9 - 4sin 3 0) 2 + (4cos 3 - 3cos0) 2 « 1. 



91. 



sin 2 cos 



(l + cos20)(l + cos0) 



:tan 



e 



92. If tan m -, and tan <£ is — , prove tan (2 + <£) - !• 

7 11 2i 

6 

93. Prove that tan- and cot- are the roots of the 

equation 

a 2 — 2fccosec0-|-l = O. 



84 PLANE TRIGONOMETRY. 



94. If tan = -, prove that 
a 



ja+± ja-b = 
\ a -b Va+b 



2eos0 



Vcos 2 e 

95. Find the values of (1) sin 9°, (2) cos 9°, (3) sin 81°, 
(4) cos 189°, (5) tan 202^°, (6) tan97£°. 

Ans. (1) | ( V3 + V5 - V5 _ V§), 

(2) \ (V3 + V5 + V5 - Vg), 

(3) sin 81°= cos 9°, 

(4) cos 189°= -cos 9°, 

(5) V2 - 1, 

(6) _(V3 + V2)(V2 + 1). 

96. If A = 200°, prove that 
. A 



(1) 2sin— = +• Vl + sinA + VI -sin A. 



A = -(l + Vl+tan*A), 
v ' 2 tan A 

97. If A lies between 270° and 360°, prove that 

(1) 2sin- = + Vl-sinA- Vl + sinA. 

A 

(2) tan - = — cot A + cosec A. 

98. If A lies between 450° and 630°, prove that 
A 



(1) 2sin— = - Vl + sinA- Vl -sin A. 



(2) 2 cos— = - Vl + sin A + Vl - sin A. 



EXAMPLES. 



85 



Prove the following statements, A, B, C being the angles 
of a triangle. 

m sin A — sin B . C , A — B 

99. — - = tan — tan 

sinA + sinB 2 2 



100. 



sin3B-sin3C_tan3A 
cos3C-cos3B~ 2 



iA1 A A , . B B , . C C 
101. sm — cos h sin— cos — f- sm - cos - 

2 2 2 2 2 2 

o A B C 

= zcos-cos-cos- 
2 2 2 



102. cos 2 - +cos 2 cos 2 - = 2 cos— cos — sin 

o ■ o o o o 



2 



iC. 

2 



A 

'2 



B . C 

— sm — 

2 2 



103. sin A cos A — sin B cos B + sin C cos C 
= 2 cos A sin B cos C. 

104. cos2A + cos2B + cos2C = — 1 — 4 cos A cos B cos C. 

105. sin 2 A - sin 2 B + sin 2 C = 2 sin A cos B sin C. 

106. tan — tan — I- tan - tan — I- tan — tan — = 1. 

2 2 2 2 2 2 

Prove the following statements when we take for sin" 1 , 
cos -1 , etc., their least positive value. 

107. sin- 1 i=cos- 1 ^ = cot~ 1 V3. 

108. 2tan-i(cos2fl) = taiiV cot>g ~ ^* X 



109. 4tan- 1 l-tan- 1 " 



7T 

239 ~ 4 



110. sin-i-H-sin-'A+siu- 1 — = -• 
5 17 85 2 



86 PLANE TRIGONOMETRY. 

111. tan- 1 V5 (2 - V3) - cot" 1 V5 (2 + V3) = cot" 1 VB. 

112. sec" 1 V3 = 2 cot" 1 V2. 

113. 2cot- 1 a: = cosec- 1 ^-^ 

114. 

115. 

116. 

117. 

118. Prove that cos (2 tan -1 x) = 

119. 

120. 





V **J — 


OUOCl 


2a; 






tan' 


.,V3 + 

V3- 


V2 
V2 


+ tan~ 


■vi= 


_3» 

= 4 


sin - 


iJL + cot- 

V5 


[q w 







cos- 1 g + 2tan- 1 i = sin- 1 |. 
65 5 5 

If 6 = sin- 1 -, and <£ = cos" 1 -, then 6 + <j> = 90° 
5 5 

1-a 2 



1 + OJ 2 



121. 
122. 
123. 

124. 
125. 



-il 



< tan- 1 i + cosec- 1 VlO = -• 
2 4 

' 2 tan- 1 cosec~ 1 -= sin -1 — 

3 3 65 



« 2tan- 1 i + cos- 1 - = -. 
2 5 2 

< sin" 1 (cos a?) + cos" 1 (sin y) + x + y = 7r. 

1 12 

tan -1 (- tan- 1 h tan" 1 — = nir. 

1 + x 1 — x xr 



i. -1 # — 1 . i 1 1 , 7T 

tan * h tan" 1 = rnr -\ — 

x 2x-l 4 



sin- 1 ^ — sin" 1 y 



= cos -1 (;n/ ± Vl — x? — y 2 + x?y 2 ). 



NATURE AND USE OF LOGARITHMS. 87 






CHAPTER IV. 

LOGABITHMS AND LOGAKITHMIC TABLES. — TBIGO- 
NOMETKIC TABLES. 

62. Nature and Use of Logarithms. — The numerical cal- 
culations which occur in Trigonometry are very much 
abbreviated by the aid of logarithms ; and thus it is neces- 
sary to explain the nature and use of logarithms, and the 
manner of calculating them. 

The logarithm of a number to a given base is the exponent 
of the power to which the base must be raised to give the 
number. 

Thus, if a x = m, x is called the " logarithm of m to the 
base a," and is usually written # = log a ra, the base being 
put as a suffix.* 

The relation between the base, logarithm, and number is 
expressed by the equation, 

(base) log = number. 

Thus, if the base of a system of logarithms is 2, then 3 
is the logarithm of the number 8, because 2 3 = 8. 

If the base be 5, then 3 is the logarithm of 125, because 
5 3 = 125. 

63. Properties of Logarithms. — The use of logarithms 
depends on the following properties which are true for all 
logarithms, whatever may be the base. 

♦From the definition it follows that (1) log a a x = x ) and conversely (2) a lo Sa"» = m. 
Taking the logarithms of both sides of the equation a x = m, we have log a a x = ar=log m. 
Conversely, taking the exponentials of both sides of x = \og a m to base a, we have 
a x = a}°Za m = 77i. a x = m and x = \og a m are thus seen to be equivalent, and to 
express the same relation between a number, w, and its logarithm, x, to base a. 






88 PLANE TRIGONOMETRY. 

(1) The logarithm of 1 is zero. 

For a = 1, whatever a may be ; therefore log 1 = 0. 

(2) The logarithm of the base of any system is unity. 
For a x = a, whatever a may be ; therefore log a = 1. 

(3) The logarithm of zero in any system whose base is 
greater than 1 is minus infinity. 

For a" 00 = — = — = ; therefore log = — <x>. 
a 00 oo 

(4) The logarithm of a product is equal to the sum of the 
logarithms of its factors. 

For let # = log a m, and 2/ = log n. 

.\ m = a x , and n = a y . 
,\ mn = a x+y . 
.\ log a mn = x + y = log a ra + log a n. 
Similarly, log a mnp = log a m + log a n + log a p, 

and so on for any number of factors. 

Thus, log 60 = log (3 x 4 x 5), 

= log3 + log4 + log5. 

(5) The logarithm of a quotient is equal to the logarithm 
of the dividend minus the logarithm of the divisor. 



For let x = log a m, and y = log a n. 

.-. m = a x , and n = a y . 

.-. ™ = a- 
n 

••• loga— =a-2/ = log a m-log a n. 
n 

Thus, log ^ = log 17 - log 5. 

5 






PROPERTIES OF LOGARITHMS. 89 

(6) The logarithm of any power of a number is equal to 
the logarithm of the number multiplied by the exponent of the 
power. 

For let x = log a ra. .*. m = a x . 

.-. m p = a px . 
. \ log a m p =px=p log a m. 

(7) The logarithm of any root of a number is equal to the 
logarithm of the number divided by the index of the root. 

For let x = log a m. .*. m = a x . 

1 X 

,\ m r = a r . 

I x i 
.-. log (m r ) = - = -log a m. 
r r 

It follows from these propositions that by means of 
logarithms, the operations of multiplication and division are 
changed into those of addition and subtraction; and the 
operations of involution and evolution are changed into those 
of multiplication and division. 

1. Suppose, for instance, it is required to find the product 
of 246 and 357; we add the logarithms of the factors, and 
the sum is the logarithm of the product : thus, 






log 10 246 = 2.39093 

log 10 357 = 2.55267 



4.94360 
which is the logarithm of 87822, the product required. 

2. If we are required to divide 371.49 by 52.376, we pro- 
ceed thus : 

log 10 371.49 = 2.56995 
log 10 52.376 = 1.71913 

0.85082 

which is the logarithm of 7.092752, the quotient required. 



90 PLANE TRIGONOMETRY. 






3. If we have to find the fourth power of 13, we proceed 

thus: 

log 10 13 = 1.11394 

4 



4.45576 
which is the logarithm of 28561, the number required. 

4. If we are to find the fifth root of 16807, we proceed 

thus: 

5 )4.22549 = log 10 16807, 

0.845098 
which is the logarithm of 7, the root required. 

5. Given log 10 2 = 0.30103 ; find log 10 128, log 10 512. 

Arts. 2.10721, 2.70927. 

6. Given log 10 3 = 0.47712 ; find log 10 81, log 10 2187. 

Ana. 1.90849, 3.33985. 

7. Givenlog 10 3; find log 10 W 3 . 0.28627. 

8. Find the logarithms to the base a of 

a , a 3 , v a, V cr, a 2 . 

9. Find the logarithms to the base 2 of 8, 64, £, .125, 
.015625, -^64. Arts. 3, 6, -1, -3, -6, 2. 

10. Find the logarithms to base 4 of 8, Vl6, V^5> 
-^.015625. Arts, f, f, -J, -1. 

Express the following logarithms in terms of log a, log&, 
and logc: 



11. logV(a 2 6 3 c) 6 . Ans. 61oga + 91og6 + 31ogc 

12. log -y/a? b 5 c 7 . |loga + flog 6 + £logc. 

13. log ^ ab ' lc ~ 2 iloga. 






SYSTEMS OF LOGARITHMS. 91 

64. Common System of Logarithms. — There are two 
systems of logarithms in use, viz., the Naperian* system 
and the common system. 

The Naperian system is used for purely theoretic investi- 
gations; its base is e = 2.7182818. 

The common system | of logarithms is the system that 
is used in all practical calculations ; its base is 10. 

By a system of logarithms to the base 10, is meant a suc- 
cession of values of x which satisfy the equation 

m = 10*, 

for all positive values of m, integral or fractional. Thus, if 
we suppose m to assume in succession every value from 
to oo, the corresponding values of x will form a system of 
logarithms, to the base 10. 

Such a system is formed by means of the series of loga- 
rithms of the natural numbers from 1 to 100000, which con- 
stitute the logarithms registered in our ordinary tables. 

Now 10° =1, .-. logl =0 

10 1 = 10, .•; log 10 =1 

10 2 = 100, .-. log 100 =2 

10 3 = 1000, .-. log 1000 = 3. 
and so on. 

Also, 10- x = T \ =.1, .-. log.l =-1; 

10~ 2 =tU =.01, ... log .01 =-2; 

10- 3 = T¥ Vo = -001, .-. log.001 = - 3. 
and so on. 

Hence, in the common system, the logarithm of any 
number between 

1 and 10 is some number between and 1 ; i.e., + 
a decimal ; 

* So called from its inventor, Baron Napier, a Scotch mathematician, 
f First introduced in 1615 by Briggs, a contemporary of Napier. 



92 PLANE TRIGONOMETRY. 






10 and 100 is some number between 1 and 2 ; i.e., 1 + 
a decimal ; 

100 and 1000 is some number between 2 and 3 ; i.e., 2 + 
a decimal ; 

1 and .1 is some number between and — 1 ; i.e., — 1 + 
a decimal ; 

.1 and .01 is some number between — 1 and — 2 ; i.e., 

— 2 + a decimal ; 

.01 and .001 is some number between — 2 and — 3 ; i.e., 

— 3+ a decimal; 

and so on. 

It thus appears that 

(1) The (common) logarithm of any number greater than 
1 is positive. 

(2) The logarithm of any positive number less than 1 is 
negative. 

(3) In general, the common logarithm of a number con- 
sists of two parts, an integral part and a decimal part. 

The integral part of a logarithm is called the characteristic 
of the logarithm, and may be either positive or negative. 

The decimal part of a logarithm is called the mantissa 
of the logarithm, and is always kept positive. 

Note. — It is convenient to keep the decimal part of the logarithms always posi- 
tive, in order that numbers consisting of the same digits in the same order may 
correspond to the same mantissa. 

It is evident from the above examples that the character- 
istic of a logarithm can always be obtained by the following 
rule: 

Rule. — The characteristic of the logarithm of a number 
greater than unity is one less than the number of digits in 
the whole number. 

The characteristic of the logarithm of a number less than 
unity is negative, and is one more than the number of ciphers 
immediately after the decimal point. 



RULES FOR THE CHARACTERISTIC. 93 

Thus, the characteristics of the logarithms of 1234, 123.4, 
1.234, .1234, .00001234, 12340, are respectively, 3, 2, 0, - 1, 
-5, 4. 

Note. — When the characteristic is negative, the minus sign is written over it to 
indicate that the characteristic alone is negative, the mantissa being always positive. 

Write down the characteristics of the common logarithms 
of the following numbers : 

1. 17601, 361.1, 4.01, 723000, 29. Ans. 4, 2, 0, 5, 1. 

2. .04, .0000612, .7963, .001201, .1. 

bAns. -2,-5, -1, -3, -1. 
3. How many digits are there in the integral part of the 
numbers whose common logarithms are respectively 3.461, 
0.30203, 5.47123, 2.67101 ? 

4. Given log 2 = 0.30103; find the number of digits in the 
integral part of 8 10 , 2 12 , 16 20 , 2 100 . Ans. 10, 4, 25, 31. 

65. Comparison of Two Systems of Logarithms. — Given 
the logarithm of a number to base a ; to find the logarithm 
of the same number to base b. 

Let m be any number whose logarithm to base b is 
required. 

Let # = log 6 ra; then b x = m. 

.-. log a (6 a; ) = log a m; or xlog a b = log a ra. 



or 



•'•* = i^b xlog ° m ' 

log>m = ^ . . . .' (1) 

Hence, to transform the logarithm of a number from 

1 



base a to base 6, we multiply it by 



94 PLANE TRIGONOMETRY. 

This constant multiplier is called the modulus of 

log. 6 J 

the system of which the base is b with reference to the system 
of which the base is a. 

If, then, a list of logarithms to some base e can be made, 
we can deduce from it a list of common logarithms by mul- 
tiplying each logarithm in the given list by the modulus 

of the common system 

lOgelO 

Putting a for m in (1), we have 

log, a m l ^ = * by (2) of Art. 63. 
log a 6 log a 6 

.-. log 6 a x log a 6 = l. 



EXAMPLES. 

1. Show how to transform logarithms with base 5 to 
logarithms with base 125. 

Let m be any number, and let x be its logarithm to base 
125. 

Then m = 125* = (5 3 ) x = 5 s *. .\ 3a = log 6 m. 

.-. aj==log 125 m = ilog 5 m. 

Thus, the logarithm of any number to base 5, divided by 
3 (i.e., by log 5 125), is the logarithm of the same number to 
the base 125. 

Otherwise by the rule given in (1). Thus, 

log 125 m = * 5 m -f— 
log 6 125 3 

Show how to transform 

2. Logarithms with base 2 to logarithms with base 8. 

Ans. Divide each logarithm by 3. 



TABLES OF LOGARITHMS. 95 

3. Logarithms with base 9 to logarithms with base 3. 

Arts. Multiply each logarithm by 2. 

4. Find log 2 8, log 5 l, log 8 2, log 7 l, log 32 128. 

Ans. 3, 0, % 0, £ 

66. Tables of Logarithms. — The common logarithms of 
all integers from 1 to 100000 have been found and registered 
in tables, which are therefore called tabular logarithms. In 
most tables they are given to six places of decimals, though 
they may be calculated to various degrees of approximation, 
such as five, six, seven, or a higher number of decimal places. 
Tables of logarithms to seven places of decimals are in 
common use for astronomical and mathematical calculations. 
The common system to base 10 is the one in practical use, 
and it has two great advantages : 

(1) From the rule (Art. 64) the characteristics can be 
written down at once, so that only the mantissse have to be 
given in the tables. 

(2) The mantissae are the same for the logarithms of all 
numbers which have the same significant digits, in the same 
order, so that it is sufficient to tabulate the mantissse of the 
logarithms of integers. 

For, since altering the position of the decimal point with- 
out changing the sequence of figures merely multiplies or 
divides the number by an integral power of 10, it follows 
that its logarithm will be increased or diminished by an 
integer; i.e., that the mantissa of the logarithm remains 
unaltered. 

In General. — If N" be any number, and p and q any 
integers, it follows that N x 10 p and N -*- 10* are numbers 
whose significant digits are the same as those of N. 

Then log (1ST x 10*) = logN +plogl0 = logN +p. (1) 
Also, log (N -*- 10«) = log ST - q log 10 = logN— q. (2) 



96 PLANE TRIGONOMETRY. 

In (1) the logarithm of N is increased by an integer, and 
in (2) it is diminished by an integer. 

That is, the same mantissa serves for the logarithms of 
all numbers, whether greater or less than unity, which have 
the same significant digits, and differ only in the position 
of the decimal point. 

This will perhaps be better understood if we take a 
particular case. 

From a table of logarithms we find the mantissa of the 
logarithm of 787 to be 895975 ; therefore, prefixing the char- 
acteristic with its appropriate sign according to the rule, 
we have 

log 787 =2.895975. 

Now log 7.87 = log Ijjl = log 787 -2 

= 0.895975. 

Also, log .0787 = log^j^) = log787 - 4 

= 2.895975. 

Also, log 78700 = log (787 x 100) = log 787 + 2 

= 4.895975. 

Note 1. — We do not write log 10 787; for so long as we are treating of logarithms 
to the particular base 10, we may omit the suffix. 

Note 2. — Sometimes in working with negative logarithms, an arithmetic artifice 
will be necessary to make the mantissa positive. For example, a result such as 
— 2.69897, in which the whole expression is negative, may be transformed by sub- 
tracting 1 from the characteristic, and adding 1 to the mantissa. Thus, 

- 2.69897 = - 3 + (1 - .69897) = 3.30103. 

Note 3. — When the characteristic of a logarithm is negative, it is often, espe- 
cially in Astronomy and Geodesy, for convenience, made positive by the addition 
of 10, which can lead to no error, if we are careful to subtract 10. 

Thus, instead of the logarithm 3.603582, we may write 7.603582 - 10. 

In calculations with negative characteristics we follow 
the rules of Algebra. 



EXAMPLES. 



97 



1. Add together 



EXAMPLES. 

2.2143 
1.3142 
5.9068 



2. From 
take 



7.4353 Ans. 

3.24569 
5.62493 



1.62076 



the 1 carried from the last subtraction in decimal places 
changes — 5 into — 4, and then — 4 subtracted from — 3 
gives 1 as a result. 






3. Multiply 2.1528 by 7. 



2.1528 

7 

13.0696 



the 1 carried from the last multiplication of the decimal 
places being added to — 14, and thus giving — 13 as a result. 

Note 4. — When a logarithm with negative characteristic has to be divided by a 
number which is not an exact divisor of the characteristic, we proceed as follows in 
order to keep the characteristic integral. Increase the characteristic numerically by 
a number which will make it exactly divisible, and prefix an equal positive number 
to the mantissa. 

4. Divide 3.7268 by 5. 

Increase the negative characteristic so that it may be 
exactly divisible by 5 ; thus 



3.7268 _ 5 + 2.7268 



= 1.5453. 



Given that log 2 =.30103, log 3 =.47712, and log 7 =.84510; 
find the values of 

5. log 6, log 42, log 16. Ans. .77815, 1.62325, 1.20412. 



98 PLANE TRIGONOMETRY. 

6. log49, log36, log63. Ans. 1.69020, 1.55630, 1.79934. 

7. log 200, log 600, log 70. 2.30103, 2.77815, 1.84510. 

8. log 60, log .03, log 1.05, log. 0000432. 

Notb. — The logarithm of 5 and its powers can always be obtained from log 2. 

Ana. 1.77815, 2.47712, .02119, 5.63548. 

9. Given log 2 =.30103; find log 128, log 125, and log 2500. 

Arts. 2.10721, 2.09691, 3.39794. 

Given the logarithms of 2, 3, and 7, as above ; find the 
logarithms of the following : 

10. 20736, 432, 98, 686, 1.728, .336. 

Ans. 4.31672, 2.63548, 1.99122, 2.83632, .23754,1.52634. 

11. V3, (.03)% (.0021)*, (.098) 3 , (.00042) 5 , (.0336)*. 
Ans. 1.65052, 1.61928, 1.46444, 4.97368, 17.11625, 1.26317. 

67. Use of Tables of Logarithms * of Numbers. — In our 

explanations of the use of tables of common logarithms we 
shall use tables of seven places of decimals.! These tables 
are arranged so as to give the mantissse of the logarithms 
of the natural members from 1 to 100000 ; i.e., of numbers 
containing from one to five digits. 

A table of logarithms of numbers correct to seven deci- 
mal places is exact for all the practical purposes of Astron- 
omy and Geodesy. For an actual measurement of any kind 
must be made with the greatest care, with the most accurate 
instruments, by the most skilful observers, if it is to attain 
to anything like the accuracy represented by seven signifi- 
cant figures. 

* The methods by which these tables are formed will be given in Chap. VIII. 

f The student should here provide himself with logarithmic and trigonometric 
tables of seven decimal places. The most convenient seven-figure tables used in this 
country are Stanley's, Vega's, Bruhns', etc. In the appendix to the Elementary 
Trigonometry are given five-figured tables, which are sufficiently near for most prac- 
tical applications. 



USE OF LOGARITHMIC TABLES. 99 

If the measure of any length is known accurately to seven figures, 
it is practically exact ; i.e., it is known to within the limits of obser- 
vation. 

If the measure of any angle is known to within the tenth part of a 
second, the greatest accuracy possible, at present, in the measurement 
of angles is reached. The tenth part of a second is about the two- 
millionth part of a radian. This degree of accuracy Is attainable only 
with the largest and best instruments, and under the most favorable 
conditions. 

On page 101 is a specimen page of Logarithmic Tables. 
It consists of the mantissae of the logarithms, correct to 
seven places of decimals, of all numbers between 62500 and 
63009. The figures of the number are those in the left 
column headed N, followed by one in larger type at the top 
of the page. The first three figures of the mantissae 795, 
796, 797 ••-, and the remaining four are in the same hori- 
zontal line with the first four figures of the number, and in 
the vertical column under the last. 

Logarithms are in general incommensurable numbers. 
Their values can therefore only be given approximately. 
Throughout all approximate calculations it is usual to take 
for the last figure which we retain, that figure which gives 
the nearest approach to the true value. When only a cer- 
tain number of decimal places is required, the general rule 
is this : Strike out the rest of the figures, and increase the last 
figure retained by 1 if the first figure struck off is 5 or greater 
than 5. 

68. To find the Logarithm of a Given Number. — When 
the given number has not more than five digits, we have 
merely to take the mantissa immediately from the table, 
and prefix the characteristic by the rule (Art. 64). 

Thus, suppose we require the logarithm of 62541. The 
table gives .7961648 as the mantissa, and the characteristic 
is 4, by the rule ; therefore 

log 62541 = 4.7961648. 
Similarly, log .006281 = 3.7980288 . . (Art. 64) 



100 PLANE TRIGONOMETRY, 

Suppose, however, that the given number has more than 
five digits. For example : 

Suppose we require to find log 62761.6. 
We find from the table 

log 62761 = 4.7976899 

log 62762 = 4.7976968 

and diff. for 1 = 0.0000069 

Thus for an increase of 1 in the number there is an in- 
crease of .0000069 in the logarithm. 

Hence, assuming that the increase of the logarithm is 
proportional to the increase of the number, then an increase 
in the number of .6 will correspond to an increase in the 
logarithm of .6 x .0000069 = .0000041, to the nearest sev- 
enth decimal place. 

Hence, log 62761 = 4.7976899 

diff. for .6 = 41 



.-. log 62761.6 = 4.7976940 

This explains the use of the column of proportional parts 
on the extreme right of the page. It will be seen that the 
difference between the logarithms of two consecutive num- 
bers is not always the same; for instance, those in the 
upper part of the page before us differ by .0000070, while 
those in the middle and the lower parts differ by .0000069 
and .0000068. Under the column with the heading 69 we 
see the difference 41 corresponding to the figure 6, which 
implies that when the difference between the logarithms of 
two consecutive members is .0000069, the increase in the 
logarithm corresponding to an increase of .6 in the number 
is .0000041 ; for .06 it is evidently .0000004, and so on. 

Note. — We assume in this method that the increase in a logarithm is propor- 
tional to the increase in the number. Although this is not strictly true, yet it is in 
most cases sufficiently exact for practical purposes. 

Had we taken a whole number or a decimal, the process would have been the 







TABLE OF LOGARITHMS. 








101 


N. 





1 


2 | 3 


4 


5 


6 


7 


8 


9 


P.P. 


6250 
51 


795 8800 


8870 


8939 


9009 


9078 


9148 


9217 


9287 


9356 9426 




9495 


9564 


9634 


9703 


9773 


9842 


9912 


9981 


00510120 


52 


796 0190,0259 


0329 


0398 


0468 


0537 


0606 


0676 


0745,0815 




53 


0884 0954 


1023 


1093 


1162 


1232 1301 


1370 


1440 1509 




51 


1579 


1648 


1718 


1787 


1857 


1926 


1995 


2065 


2134 


2204 




55 


2273 


2343 


2412 


2481 


2551 


2620 


2690 


2759 


2829 


2898 




56 


2967 


3037 


3106 


3176 


3245 


3314 


3384 


3453 


3523 


3592 




70 


57 


3662 


3731 


3800 


3870 


3939 


4009 


4078 


4147 


4217 


4286 


1 


7.0 


58 


4356 


4425 


4494 


4564 


4633 


4703 


4772 


4841 


4911 


4980 


2 


14.0 


59 

6260 

61 


5050 


5119 


5188 


5258 


5327 


5396 


5466 


5535 


5605 5674 


3 

4 

5 


21.0 
28.0 
35.0 


796 5743 


5813 


5882 


5951 


6021 


6090 


6160 


6229 


6298 6368 


6437 


6506 6576 


6645 


6714 


6784 


6853 


6923 


6992 


7061 


62 


7131 


7200 7269 


7339 


7408 


7477 


7547 


7616 


7685 


7755 


6 


42.0 


63 


7824 


7893 7963 


8032 


8101 


8171 


8240 


8309 


8379 


8448 


7 


49.0 


64 


8517 


8587|8656 


8725 


8795 


8864 


8933 


9003 


9072 


9141 


8 


56.0 


65 


9211 


9280J9349 


9419 


9488 


9557 


9627 


9696 


9765 


9835 


9 


63.0 


66 


9904 


9973 0043 


0112 


0181 


0250 0320 


0389 


0458 


0528 




67 


797 059 7 


0666|0736 


0805 


0874 


0943 


1013 


1082 


1151 


1221 




68 


1290 


1359 1428 


1498 


1567 


1636 


1706 


1775 


1844 


1913 




69 

6270 

71 


1983 


205212121 


2191 


2260 


2329 


2398 


2468 


2537 


2606 




797 2675 


27452814 


2883 


2952 


3022 


3091 


3160 


3229 3299 


3368 


3437 


3507 


3576 


3645 


3714 


3784 


3853 


3922 


3991 




69 


72 


4060 


4130 


4199 


4268 


4337 


4407 


4476 


4545 


4614 


4684 


1 


6.9 


73 


4753 


4822 


4891 


4961 


5030 


5099 


5168 


5237 5307 


5376 


2 


13.8 


74 


5445 


5514 5584 


5653 


5722 


5 791 


5860 


5930 5999 


6068 


3 


20.7 


75 


6137 6207 62766345 


6414 


6483 


6553 


66226691 


6760 


4 


27.6 


76 


682916899 6968 7037 


7106 


7175 


7245 


7314 7383 


7452 


5 


34.5 


77 


7521 


7590 


7660 


77297798 


7867 


7936 


8006 


8075 


8144 


6 


41.4 


78 


8213 


8282 


8351 


84218490 


8559 


8628 8697 


8766 


8836 


7 


48.3 


79 
6280 

81 


8905 


8974 


9043 


91129181 


9251 


93209389 


9458 


9527 


8 
9 


55.2 
62.1 


797 9596 9666 


9735 


9804 9873 


9942 


00110080 0150 


0219 


798 028810357 0426 0495 0565 


0634 


0703 0772 0841 


0910 




82 


0979 


1048 


111811871256 


1325 


1394 


1463 


1532 


1601 




83 


1671 


1740 


1809 


1878 1947 


2016 


2085 


2154 


2224 


2293 




84 


2362 


2431 


2500 


25692638 


2707 


2776 


2846 


2915 


2984 




85 


3053 


3122 


3191 


32603329 


3398 


3467 


3536 


3606 


3675 




86 


3744 


3813 


3882 


39514020 


40894158 


4227 


4296 


4366 




68 


87 


4435 


4504 


4573 


46424711 


4780 4849 


4918 


4987 


5056 


1 


6.8 


88 


5125 


5194 


5263 


53335402 


5471 


5540 


5609 


5678 


5747 


2 


13.6 


89 

6290 

91 


5816 5885 5954 


6023 6092 


6161 


6230 


6299 6368 


6437 


3 

4 

5 


20.4 
27.2 
34.0 


798 6506 6575 6645 


6714:6783 


6852 


6921 


6990 


7059 


7128 


7197 7266 


7335 


7404 7473 


7542 


7611 


7680 7749 


7818 


92 


7887 


7956 


8025 


80948163 


8232 8301 


83708439 


8508 


6 


40.8 


93 


8577 


8646 


8715 


8784 8853 


8922 ! 8991 


9060|9129 


9198 


7 


47.6 


94 


9267 


9336 


9405,9474 9543 


9612J9681 9750 9819 


9888 


8 


54.4 


95 


9957 


0026 0095 0164 0233 


0302 0371 !0440 ! 0509 


0578 


9 


61.2 


96 


799 0647 


0716 0785 0854 0923 


0992 


1061 11301199 


1268 




97 


1337 


1406 14751544 1613 


1682 


1751 18201889 


1958 




98 


202712096:2164 2233 2302 


23/1 


2440(2509,2578 


2647 




99 
6300 


2716 2785 2854 2923 2992 


3061 


31301319913268 


3337 




799 3405 3474 


3543 3612 3681 


3750 3819 3888 


3957 


4026 


N. 


j 1 


2 | 3 ! 4 


5 ! 6 j 7 


8 


9 


P.P. 



102 PLANE TRIGONOMETRY. 

Thus, suppose we require to find log 627616 and log .627616. The mantissa io 
exactly the same as before (Art. 66), and the only difference to be made in the final 
result is to change the characteristic according to rule (Art. 64). 

Thus log 627616 = 5.7976942, 

and log .627616 =1.7976942. 

69. To find the Number corresponding to a Given Loga- 
rithm. — If the decimal part of the logarithm is found ex- 
actly in the table, we can take out the corresponding 
number, and put the decimal point in the number, in the 
place indicated by the characteristic. 

Thus if we have to find the number whose logarithm is 
2.7982915, we look in the table for the mantissa .7982915, 
and we find it set down opposite the number 62848 : and 
as the characteristic is 2, there must be one cipher before 
the first significant figure (Art. 64). 

Hence 2.7982915 is the logarithm of .062848. 

Next, suppose that the decimal part of the logarithm is 
not found exactly in the table. For example, suppose we 
have to find the number whose logarithm is 2.7974453. 

We find from the table 

log 62726 = 4.7974476 
log 62725 = 4.7974407 

diff.forl = .0000069 

Thus for a difference of 1 in the numbers there is a 
difference of .0000069 in the logarithms. The excess of 
the given mantissa above .7974407 is (.7974453 — .7974407) 
or .0000046. 

Hence, assuming that the increase of the number is 
proportional to the increase of the logarithm, we have 
.0000069 : .0000046 : : 1 : number to be added to 627.25. 

,. number to be added = ' 0m6 = *g = .667 G9)46.0(.666 

.0000069 69 41 4 

. \ log 62725.667 = 4.7974453, ~460 

and .-. log627.25667 = 2.7974453; 414 

therefore number required is 627.25667. 460 



ARITHMETIC COMPLEMENT. 103 

We might have saved the labor of dividing 46 by 69, by 
using the table of proportional parts as follows : 



given mantissa = 


.7974453 


mantissa of 62725 = 


.7974407 


diff. of mantissae = 


46 


rtional part for .6 = 


41.4 




4.6 


" " " .06 == 


4.14 




.46 


" " « .006 = 


.414 



and so on. 
.-. number = 627.256666-... 

69 a. Arithmetic Complement. — By the arithmetic com- 
plement of the logarithm of a number, or, briefly, the 
cologarithm of the number, is meant the remainder found 
by subtracting the logarithm from 10. To subtract one 
logarithm from another is the same as to add the co- 
logarithm and then subtract 10 from the result. 

Thus, a - b = a + (10 - b) - 10, 

where a and b are logarithms, and 10 — 6 is the arithmetic 
complement of b. 

When one logarithm is to be subtracted from the sum of 
several others, it is more convenient to add its cologarithm 
to the sum, and reject 10. The advantage of using the 
cologarithm is that it enables us to exhibit the work in a 
more compact form. 

The cologarithm is easily taken from the table mentally 
by subtracting the last significant figure on the right from 
10, and all the others from 9. 



104 



PLANE TRIGONOMETRY. 



1. 


Given 


find 




2. 


Given 


find 




3. 


Given 


find 




4. 


Given 


find 




5. 


Given 


find the number 


6. 


Given 


find the number 


7. 


Given 


find the number 


8. 


Given 


find the number 



Ans. 4.7201799. 



EXAMPLES. 

log 52502 =4.7201758, 
log 52503 =4.7201841; 
log 52502.5. 

log 3.0042 = 0.4777288, 
log 3.0043 = 0.4777433; 
log 300.425. 2.4777360. 

log 7.6543 = 0.8839055, 
log 7.6544 = 0.8839112; 
log 7.65432. 

log 6.4371 = 0.8086903, 
log 6.4372 = 0.8086970; 
log 6437125, 6.8086920. 

log 12954 =4.1124039, 
log 12955 =4.1124374; 
whose logarithm is 4.1124307. 12954.8. 

log 60195 =4.7795532, 
log 60196 =4.7795604; 
whose logarithm is 2.7795561. 601.95403. 

log 3. 7040 = .5686710, 
log 3.7041 = .5686827; 
whose logarithm is .5686760. 3.70404. 

log 2.4928 = .3966874, 

log 2.4929 = .3967049; 

whose logarithm is 6.3966938. 2492837. 



. 









NATURAL TRIGONOMETRIC FUNCTIONS. 105 

9. Given log 32642 = 4.5137768, 

log 32643 =4.5137901; 
find log 32642.5. Ans. 4.5137835. 

10. Find the logarithm of 62654326. 7.7969510. 
Use specimen page. 

11. Find the number whose logarithm is 4.7989672. 

Ans. 62945.876. 

70. Use of Trigonometric Tables. — Trigonometric Tables 
are of two kinds, — Tables of Natural Trigonometric Functions 
and Tables of Logarithmic Trigonometric Functions. As the 
greater part of the computations of Trigonometry is carried 
on by logarithms, the latter tables are by far the most use- 
ful. 

We have explained in Art. 27 how to find the actual 
numerical values of certain trigonometric functions, exactly 
or approximately. 

Thus, sin30° = r ; that is, .5 exactly. 

Also, tan 60° = V3 ; that is, 1.73205 approximately. 

A table of natural trigonometric functions gives their 
approximate numerical values for angles at regular intervals 
in the first quadrant. In some tables the angles succeed 
each other at intervals of 1", in others, at intervals of 10", 
but in ordinary tables at intervals of V : and the values of 
the functions are given correct to five, six, and seven places. 
The functions of intermediate angles can be found by the 
principle of proportional parts as applied in the table of 
logarithms of numbers (Arts. 68 and 69). 

It is sufficient to have tables which give the functions 
of angles only in the first quadrant, since the functions of 
all angles of whatever size can be reduced to functions 
of angles less than 90° (Art. 35). 



106 PLANE TRIGONOMETRY. 






71. Use of Tables of Natural Trigonometric Functions. — 

These tables, which consist of the actual numerical values 
of the trigonometric functions, are commonly called tables 
of natural sines, cosines, etc., so as to distinguish them from 
the tables of the logarithms of the sines, cosines, etc. 

We shall now explain, first, how to determine the value 
of a function that lies between the functions of two con- 
secutive angles given in the tables ; and secondly, how to 
determine the angle to which a given ratio corresponds. 

72. To find the Sine of a Given Angle. 

Find the sine of 25° 14' 20", having given from the table 

sin 25° 15' = . 4265687 
sin 25° 14'=. 4263056 



diff. for 1' = .0002631 

Let d = diff . for 20" ; and assuming that an increase in 
the angle is proportional to an increase in the sine, we have 

60 : 20 : : .0002631 : cl 

,.d = 2Qx ' Q002631 = . 0000877. 
60 

.-. sin 25° 14' 20" = .4263056 + .0000877 

= .4263933. 

Note. — We assumed here that an increase in the angle is proportional to the 
increase in the corresponding sine, which is sufficiently exact for practical purposes, 
with certain exceptions. 

73. To find the Cosine of a Given Angle. 

Find the cosine of 44° 35' 25", having given from the table 

cos 44° 35' = .7122303 
cos 44° 36' = .7120260 



diff. for 1' = .0002043 

observing that the cosine decreases as the angle increases 
from 0° to 90°. 



EXAMPLES. 



107 



Let d — decrease of cosine for 25" ; then 



60:25:: .0002043 : d. 



.-. d = — x .0002043 = .0000851. 
60 

.-. cos44° 35' 25" = .7122303 - .0000851 

= 7121452. 

Similarly, we may find the values of the other trigono- 
metric functions, remembering that, in the first quadrant, 
the tangent and secant increase and the cotangent and 
cosecant decrease, as the angle increases. 



1. 


Given 


find 




2. 


Given 


find 
3. 


Given 


find 
4. 


Given 


find 




5. 


Given 



find 



EXAMPLES. 

sin 44° 35'= .7019459, 
sin 44° 36'= .7021531; 
sin 44° 35' 25".. 

sin 42° 15'= .6723668, 
sin 42° 16'= .6725821; 
sin 42° 15' 16". 

sin 43° 23'= .6863761, 
sin 43° 22'= .6866647; 
sin 43° 22' 50". 

sin 31° 6' =.5165333, 
sin 31° 7' =.5167824; 
sin 31° 6' 25". 

cos 74° 45'= .4265687, 
cos 74° 46'= .4263056; 
cos 74° 45' 40". 



Arts. .7020322. 



.6724242. 



.6868408. 



.5166371. 



.4263933. 



108 




PLANE TBIGONOMETR 


6. 


Given 


cos 41° 13'= .7522233, 
cos 41° 14'= .7520316; 


find 




cos 41° 13' 26". 


7. 


Given 


cos 47° 38'= .6738727, 
cos 47° 39'= .6736577; 


find 




cos47°38'30". 






Ans. .7521403. 



.6737652. 

74. To find the Angle whose Sine is Given. 

Find the angle whose sine is .5082784, having given from 

sin 30° 33' = .5082901 
sin 30° 32' = .5080396 

diff. for 1'=. 0002505 

given sine = .5082784 
sin 30° 32' = .5080396 



the table 



diff. = .0002388 

Let d = diff. between 30° 32' and required angle ; then 
.0002505 : .0002388 : : 60 : d. 



d = 



2388 x 60 6552 



2505 167 

= 57.2 nearly. 
.-. required angle = 30° 32' 57".2. 

75. To find the Angle whose Cosine is Given. 

Find the angle whose cosine is .4043281, having given 

cos 66° 9' =.4043436 
cos 66° 10' = .4040775 



from the table 



diff. for 1' = .0002661 

cos 66° 9' = .4043436 
given cosine = .4043281 

diff. = .0000155 



EXAMPLES. 109 

Let d = cliff, between 66° 9' and required angle ; then 
.0002661 : .0000155 : : 60 : d. 

... d= 1BBxG0 = 3.5. 
2661 

Required angle is greater than 66° 9' because its cosine is 
less than cos 66° 9'. 

.-. required angle = 66° 9' 3".5. 



EXAMPLES. 

1. Given sin 44° 12' = .6971651, 

sin 44° 11' =.6969565; 
find the angle whose sine is .6970886. Ans. 44° 11' 38". 

2. Given sin 48° 47' = .7522233, 

sin 48° 46' = .7520316; 
find the angle whose sine is .752140. 48° 46' 34". 

3. Given sin 24° 11' = .4096577, 

sin 24° 12' = .4099230; 
find the angle whose sine is .4097559. 24° 11' 22".2. 

4. Given cos 32° 31' = .8432351, 

cos 32° 32' = .8430787; 
find the angle whose cosine is .8432. 32° 31' 13".5. 

5. Given cos 44° 11' = .7171134, 

cos 44° 12' = .7169106; 
find the angle whose cosine is .7169848. 44° 11' 38". 

6. Given cos 70° 32' = .3332584, 

cos 70° 31' =.3335326; 
find the angle whose cosine is .3333333. 70° 31' 43".6. 



110 PLANE TRIGONOMETRY. 






76. Use of Tables of Logarithmic Trigonometric Func- 
tions. — Since the sines, cosines, tangents, etc., of angles 
are numbers, we may use the logarithms of these numbers 
in numerical calculations in which trigonometric functions 
are involved; and these logarithms are in practice much 
more useful than the numbers themselves, as with their 
assistance we are able to abbreviate greatly our calcula- 
tions ; this is especially the case, as we shall see hereafter, 
in the solution of triangles. In order to avoid the trouble 
of referring twice to tables — first to the table of natural 
functions for the value of the function, and then to a table 
of logarithms for the logarithm of that function — the log- 
arithms of the trigonometric functions have been calculated 
and arranged in tables, forming tables of the logarithms of 
the sines, logarithms of the cosines, etc. ; these tables are 
called tables of logarithmic sines, logarithmic cosines, etc. 

Since the sines and cosines of all angles and the tangents 
of angles less than 45° are less than unity, the logarithms of 
these functions are negative. To avoid the inconvenience of 
using negative characteristics, 10 is added to the logarithms 
of all the functions before they are entered in the table. 
The logarithms so increased are called the tabular logarithms 
of the sine, cosine, etc. Thus, the tabular logarithmic sine 
of 30° is 

10 + log sin 30° = 10 + logi = 10 - log 2 = 9.6989700. 

z 

In calculations we have to remember and allow for this 
increase of the true logarithms. When the value of any 
one of the tabular logarithms is given, we must take away 
10 from it to obtain the true value of the logarithm. 

Thus in the tables we find 

log sin 31° 15' = 9.7149776. 

Therefore the true value of the logarithm of the sine of 
31° 15' is 9.7149776 - 10 = 1.7149776. 

Similarly with the logarithms of other functions. 



TABLES OF LOGARITHMIC FUNCTIONS. Ill 

Note. — English authors usually denote these tabular logarithms by the letter L, 
Thus, L sin A denotes the tabular logarithm of the sine of A. 

French authors use the logarithms of the tables diminished by 10. Thus, 

log sin A = 1.8598213, instead of 9.8598213. 

The Tables contain the tabular logs of the functions of all 
angles in the first quadrant at intervals of 1' ; and from 
these the logarithmic functions of all other angles can be 
found.* 

Since every angle between 45° and 90° is the complement 
of another angle between 45° and 0°, every sine, tangent, 
etc., of an angle less than 45° is the cosine, cotangent, etc., 
of another angle greater than 45° (Art. 16). Hence the 
degrees at the top of the tables are generally marked from 
0° to 45°, and those at the bottom from 45° to 90°, while 
the minutes are marked both in the first column at the left, 
and in the last column at the right. Every number there- 
fore in each column, except those marked diff., stands for 
two functions — the one named at the top of the column, 
and the complemental function named at the bottom of the 
column. In looking for a function of an angle, if it be less 
than 45°, the degrees are found at the top, and the minutes 
at the left-hand side. If greater than 45°, the degrees are 
found at the foot, and the minutes at the right-hand side. 

On page 113 is a specimen page of Mathematical Tables. 
It gives the tabular logarithmic functions of all angles between 
38° and 39°, and also of those between 51° and 52°, both 
inclusive, at intervals of 1'. The names of the functions 
for 38° are printed at the top of the page, and those for 51° 
at the foot. The column of minutes for 38° is on the left, 
that for 51° is on the right. 

Thus we find 

log sin 38° 29' =9.7939907. 
log cos 38° 45' =9.8920303. 
log tan 51° 18' = 10.0962856. 

* Many tables are calculated for angles at intervals of 10*. 



112 PLANE TRIGONOMETRY. 

77. To find the Logarithmic Sine of a Given Angle. 

Find log sin 38° 52' 46". 
We have from page 113 

log sin 38° 53' = 9.7977775 

log sin 38° 52' = 9.7976208 

diff. for V = .0001567 

Let d = diff. for 46", and assuming that the change in 
the log sine is proportional to the change in the angle, we 

have 

60 : 46 : : .0001567 : d. 

,. d = 4g X. 0001567 =a000120L 
60 

.-. log sin 38° 52' 46"= 9.7976208 + .0001201 

= 9.7977409. 

78. To find the Logarithmic Cosine of a Given Angle. 

Find log cos 83° 27' 23", having given from the table 

log cos 83° 27'= 9.0571723 
log cos 83° 28'= 9.0560706 

diff. for 1'= .0011017 

Let d = decrease of log cosine for 23"; then 

60 : 23 : : .0011017 : d. 

, 23 x .0011017 AAAylooo , 

.-. d = = .0004223, nearly. 

60 J 

.\ log cos 83° 27' 23"= 9.0571723 - .0004223 

= 9.0567500. 

EXAMPLES. 

1. Given log sin 6° 33'= 9.0571723, 
log sin 6° 32'= 9.0560706; 
find log sin 6° 32' 37". Arts. 9.05675. 



38 Deg. 



TABLE OF LOGARITHMS. 



113 



Sine, 



Diff. 



9.7893420 
9.7895036 
9.7896652 
9.7898266 

4 j 9.7899880 

5 9-790I493 



6 9.7903104 

7 97904715 
; l 9.7906325 

97907933 
97909541 



9.7911148 

9.7912754 
97914359 
97915903 
9.7917566 



9.7919168 
9.7920769 
9.7922369 
9.7923968 
9.7925566 
9.7927163 
9.7928760 

97930355 
9.7931949 

97933543 



26 I 97935 135 
9.7936727 

9793 8 3 x 7 
97939907 
9.7941496 



97943o83 
9.7944670 
9.7946256 
9.7947841 
9.7949425 



9.7951008 
97952590 
9.7954171 

9795575 1 

40 1 97957330 

41 9.7958909 

42 9.7960486 

43 I 9.7962062 

44 97963638 

45 97965212 

46 9.7966786 

47 97968359 

48 9.7969930 

49 9797i5oi 

50 97973071 

51 

52 
53 
54 



97974640 
9.7976208 
9.7977775 

97979341 
55 97980906 



56 9.7982470 

57 97984034 

58 97985596 

59 979 8 7i58 

60 I 9.7988718 



Tang. 



Diff. 



Cotang. 



Diff. : Cosine. 



616 
616 
614 
614 
613 
611 
611 
610 
608 
608 
607 
606 
605 
604 
603 
602 
601 
600 
599 
598 
597 
597 
595 
594 
594 
592 

592 
590 
59o I 
589 
587 

587 
586 
585 
584 
583 
582 
58i 
580 

579 
579 
577 
576 
S 76 
574 
574 

573 
57i 
57i I 
570 

569 
568 j 
567 
566 ^ 
565; 
564 

5 6 4 
562 
562 
560 



9.8928098 
9.8930702 
9-89333o6 
9.8935909 
9.8938511 
9.8941114 



9.8943715 
9.8946317 
9.8948918 
9.8951519 
9.89541 19 



9.8956719 

9-89593 J 9 
9.8961918 
9.8964517 
9.8967116 



9.8969714 
9.8972312 
9.8974910 
9.8977507 
9.8980104 



9.8982700 
9.8985296 
9.8987892 
9.8990487 



9.8995677 
9.8998271 
9.9000865 
9.9003459 
9.9006052 

9.9008645 
9.901 1 237 
9.9013830 
9.9016422 
9.9019013 
9.9021604 
9.9024195 
9.9026786 
9.9029376 
9.9031966 

9-9034555 
9.9037144 

9-9039733 
9.9042321 
9.9044910 



9.9047497 
9.9050085 
9.9052672 
9-9055259 
9.9057845 



9.9060431 
9.9063017 
9.9065603 
9.9068188 
9.90 70773 

9-9073357 
9-907594I 
9.9078525 
9.908 1 109 
9.9083692 



2604 
2604 
2603 
2602 
2603 
2601 
2602 
2601 
2601 
2600 
2600 
2600 
2599 
2599 
2599 
2598 
2598 
2598 
2597 
2597 
2596 
2596 
2596 
2595 
2595 
2595 
2594 
2594 
2594 
2593 
2593 
2592 
2593 
2592 
2591 
2591 
2591 
2591 
2590 
2590 
2589 

2589 
2589 
2588 
2589 
2587 
2588 
2587 
2587 
2586 
2586 
2586 
2586 
2585 
2585 
2584 
2584 
2584 
2584 
2583 



Cosine. Diff. Cotang, 



0.107 1902 
0.1069298 
0.1066694 
0.106409 1 
0.106 1489 
0.1058886 
0.1056285 
0.1053683 
0.1051082 
0.104848 1 
0.1045881 



0.104328 1 
0.104068 1 
0.1038082 
0.1035483 
0.1032S84 



0.1030286 
0.1027688 
0.1025090 
0.1022493 
0.1019896 



0.1017300 
0.1014704 
0.1012108 
0.1009513 
0.1006918 



0.1004323 
0.1001729 
0.0999135 
0.0996541 
0.0993948 



0.0991355 
0.0988763 
0.0986170 
0.0983578 
0.0980987 



0.0965445 
0.0962856 
0.0960267 
0.0957679 
0.0955090 



0.0952503 
0.0949915 
0.0947328 
0.0944741 
0.0942155 



0.0939569 
0.0936983 
0.0934397 
0.0931812 
0.0929227 



0.0926643 
0.0924059 
0.0921475 
0.0918891 
0.0916308 



0.0978396 
0.0975805 
0.0973214 
0.0970624 
0.0968034 



Diff. 



Tang. 



987 



989 
99o 
990 

991 
992 
992 
992 
993 
994 
995 
995 
995 
997 
996 



1000 

999 
001 
001 
001 
003 
002 
004 
004 
004 
005 
006 
007 
007 
007 
008 
009 
010 
010 
010 
on 
012 
013 
013 
013 
014 

015 
016 
016 
016 
018 
017 
019 
019 
020 
020 
021 
021 
022 
023 



9.8965321 

9-8964334 
9.8963346 
9.8962358 I 57 

9.8961369 \ 56 
9.8960379 



58 



9.8959389 
9.8958398 
9.8957406 
9.8956414 
9.8955422 



Diff. 



9.8954429 

9.8953435 
9.8952440 
9.8951445 
9-8950450 
9.8949453 
9.8948457 
9.8947459 
9.8946461 
9.8945463 



9.8944463 
9.8943464 
9.8942463 
9.8941462 
9.8940461 



9.8939458 
9.8938456 
9-8937452 
9.8936448 
9.8935444 



9-8934439 
9.8933433 
9.8932426 
9.8931419 
9.8930412 



9.8929404 
9.8928395 
9.8927385 

9.8926375 
9.8925365 



9.8924354 
9.8923342 
9.8922329 
9.8921316 
9.8920303 



9.891. 

9.8913191 

9.8912172 

9.8911153 

9.8910133 



9.8909113 
9.8908092 
9.8907071 
9.8906049 
9.8905026 



9.8919289 
9.8918274 
9.8917258 
9.8916242 n 
9.8915226 



55 
54 
53 
52 

5i 
50 

49 
48 

47 
46 

45 
44 
43 
42 
41 
40 

39 
38 
37 
36 
35 
34 
33 
32 
3i 
30 

29 
28 

27 
26 

25 
24 
23 
22 
21 
20 

19 
18 

17 

16 

15 
14 
13 
12 



Sine. 
51 Deg. 



114 




PLANE TBIGONOMETBY. 




2. 


Given 


log sin 55° 33'= 9.9162539, 
log sin 55° 34'= 9.9163406 ; 




find 




log sin 55° 33' 54". Ana. 


9.9163319. 


3. 


Given 


log cos 37° 28'= 9.8996604, 
log cos 37° 29'= 9.8995636 ; 




find 




log cos 37° 28' 36". 


9.8996023. 


4. 


Given 


log cos 44° 35' 20"= 9.8525789, 
log cos 44° 35' 30"= 9.8525582 ; 




find 




logcos44°35'25".7. 


9.8525671. 


See foot-note of Art. 76. 




5. 


Given 


log cos 55° 11'= 9.7565999, 
log cos 55° 12'= 9.7564182 ; 




find 




log cos 55° 11' 12". 


9.7565636. 


6. 


Given 


log tan 27° 13'= 9.7112148, 
log tan 27° 14'= 9.7115254 ; 




find 




logtan27°13'45". 


9.7114477. 






79. To find the Angle whose Logarithmic Sine is Given. 

Find the angle whose log sine is 8.8785940, having given 
from the table 

log sin 4° 21'= 8.8799493 
log sin 4° 20' = 8.8782854 

diff. forl'= .0016639 

given log sine = 8.8785940 
log sin 4° 20'= 8.8782854 

diff. = .0003086 
Let d = diff. between 4° 20' and required angle ; then 
.0016639 : .0003086 : : 60 : d. 
3086 x 60 



\ d = % 

16639 

,\ required angle : 



: 24, nearly. 
: 4° 20' 24". 



EXAMPLES. 115 

80. To find the Angle whose Logarithmic Cosine is Given. 

Find the angle whose log cosine is 9.8934342. 
We have from page 113 

log cos 38° 31'= 9.8934439 
log cos 38° 32'= 9.8933433 

diff. for 1'= .0001006 

log cos 38° 31'= 9.8934439 
given log cosine = 9.8934342 

diff. = .0000097 
Let d = diff. between 38° 31' and required angle ; then 
.0001006 : .0000097 : : 60 : d. 

,, d = -0000097 x6Q = 97x^0 , 
.0001006 1006 

.-. required angle = 38° 31' 5".8. 

>s"ote. — In using both the tables of the natural sines, cosines, etc., and the tables 
of the logarithmic sines, cosines, etc., the student will remember that, in the first 
quadrant, as the angle increases, the sine, tangent, and secant increase, but the 
cosine, cotangent, and cosecant decrease. 

EXAMPLES. 

1. Given log sin 14° 24'= 9.3956581, 

log sin 14° 25'= 9.3961499; 
find the angle whose log sine is 9.3959449. Ans. 14° 24' 35". 

2. Given log sin 71° 40'= 9.9773772, 

I log sin 71° 41'= 9.9774191 ; 

find the angle whose log sine is 9.9773897. 71° 40' 18". 



3. Given log cos 28° 17'= 9.9447862, 
log cos 28° 16'= 9.9448541; 
ind the angle whose log cosine is 9.9448230. 28° 16' 27".5. 



116 PLANE TRIGONOMETRY. 

4. Given log cos 80° 53'= 9.1998793, 

log cos 80° 52' 50"= 9.2000105 ; 
find the angle whose log cosine is 9.2000000. 

Ans. 80° 52' 51". 

5. Given log tan 35° 4'= 9.8463018, 

log tan 35° 5'= 9.8465705; 
find the angle whose log tangent is 9.8464028. 35° 4' 23". 

6. Given log sin 44° 35' 30"= 9.8463678, 

log sin 44° 35' 20"= 9.8463464 ; 
find the angle whose log sine is 9.8463586. 44° 35' 25".7. 

7. Find the angle by page 113 whose log tangent is 
10.1018542. Ans. 51° 39' 28". 7. 

81. Angles near the Limits of the Quadrant. — It was 

assumed in Arts. 72-80 that, in general, the differences of 
the trigonometric functions, both natural and logarithmic, 
are approximately proportional to the differences of their 
corresponding angles, with certain exceptions. The excep- 
tional cases are as follows : 

(1) Natural functions. — For the sine the differences are 
insensible for angles near 90°; for the cosine they are in- 
sensible for angles near 0°. For the tangent the differences 
are irregular for angles near 90°; for the cotangent they are 
irregular for angles near 0°. 

(2) Logarithmic functions. — The principle of propor- 
tional parts fails both for angles near 0° and angles near 
90°. For the log sine and the log cosecant the differences are 
irregular for angles near 0°, and insensible for angles near 90°. 
For the log cosine and the log secant the differences are in- 
sensible for angles near 0°, and irregular for angles near 90°. 
For the log tangent and the log cotangent the differences are 
irregular for angles near 0° and angles near 90°. 



i 



EXAMPLES. 117 

It follows, therefore, that angles near 0° and angles near 
90° cannot be found with exactness from their log trigono- 
metric functions. These difficulties may be met in three 
ways. 

(1) For an angle near 0° use the principle that the sines 
and tangents of small angles are approximately proportional 
to the angles themselves. (See Art. 130.) 

(2) For an angle near 90° use the half angle (Art. 99). 

(3) In using the proportional parts, find two, three, or 
j ^ore orders of differences (Alg., Art. 197). 

Special tables are employed for angles near the limits of 
the quadrant. 

EXAMPLES. 

1. Given log 10 7 = .8450980, find log 10 343, log 10 2401, and 
log 10 16.807. Ans. 2.5352940, 3.3803920, 1.2254900. 

2. Find the logarithms to the base 3 of 9, 81, -J-, ^j, .1, $.. 

Ans. 2, 4, - 1, -3, - 2, - 4. 

3. Find the value of log 2 8, log 2 .5, log 3 243, log 5 (.04), 
log 10 1000, log 10 .001. Ans. 3, - 1, 5, - 2, 3, - 3. 

4. Find the value of log a a J , log 6 -\^, log 8 2, log^ 3, log 100 10. 

/i^o 4 2 111 

XL/Lb. -3, -3, 3, 3, 2 « 

Given log 10 2 = .3010300, log 10 3 = .4771213, and log 10 7 = 
.8450980, find the values of the following : 

5. log 10 35, log 10 150, log 10 .2. 

Ans. 1.544068, 2.1760913, 1.30103. 

6. log 10 3.5, log 10 7.29, log 10 . 081. 

Ans. .5440680, .8627278, 2.9084852. 

7. log 10 fc log 10 3 5 , logjo^. 
Ans. .3679767, 2.3856065, .0780278. 



118 PLANE TRIGONOMETRY. 

8. Write down the integral part of the common loga- 
rithms of 7963, .1, 2.61, 79.6341, 1.0006, .00000079. 

Ans. 3, - 1, 0, 1, 0, - 7. 

9. Give the position of the first significant figure in the 
numbers whose logarithms are 

2.4612310, 1.2793400, 6.1763241. 

10. Give the position of the first significant figure in 
the numbers whose logarithms are 4.2990713, .3040595, 
2.5860244, 3.1760913, 1.3180633, .4980347. 

Ans. ten thousands, units, hundreds, 3rd dec. pi., 1st 
dec. pi., units. 

11. Given log 7 = .8450980, find the number of digits in 
the integral part of 7 10 , 49 6 , 343"*", (-V ) 20 , (4.9) M , (3.43) 10 . 

Ans. 9, 11, 85, 4, 9, 6. 

12. Find the position of the first significant figure in the 
numerical value of 20 7 , (.02) 7 ; (.007) 2 , (3.43)tV (.0343) 8 , 
(.0343) A 

Ans. tenth integral pi., 12th dec. pi., 5th dec. pi., units, 
12th dec. pi., 1st dec. pi. 

Show how to transform 

13. Common logarithms to logarithms with base 2. 

Ans. Divide each logarithm by .30103. 

14. Logarithms with base 3 to common logarithms. 

Ans. Multiply each log by .4771213. 

15. Given log 10 2 = .3010300, find log 2 10. 3.32190. 

16. Given log 10 7 = .8450980, find log 7 10. 1.183. 

17. Given log 10 2 = .3010300, find log 8 10. 1.10730. 

18. The mantissa of the log of 85762 is 9332949; find 
(1) the log of a/.0085762, and (2) the number of figures in 
(85762) n , when it is multiplied out. 

Ans. (1) 1.8121177, (2) 55. 



EXAMPLES. 



119 



19. What are the characteristics of the logarithms of 
3742 to the bases 3, 6, 10, and 12 respectively ? 

Ans. 7, 4, 3, 3. 

20. Prove that 7 log |£ + 6 log f + 5 log f + log f f = log 3. 

1 






Ans. 2 + 



21. Given log 10 7, find log 7 490. 

22. From 5.3429 take 3.6284. 

23. Divide 13.2615 by 8. 

24 Prove that 6 log f + 4 log T \ + 2 log - 2 / = 0. 
25. Find log ^297 Vil}* to the base 3VH 

Given log 2 = .3010300, log 3 = .4771213. 



log 10 7 
3.7145. 

2.4076. 



1.8. 



26. Find log 216, 6480, 5400, f. 

Ans. 2.3344539, 3.8115752, 3.7323939, 1.6478174. 

27. Find log .03, 6 - *, (S*) - *. 

Ans. 2.4771213, 1.7406162, 1.6365006. 

28. Find log .18, log 2.4, lbg^. 

Ans. 1.2552726, .3802113, 1.2730013. 



.1136971, 1.45154. 



29. Find log (6.25)?, log4V.005. 

log 56321 = 4.7506704, 
log 56322 = 4.7506781; 
log 5632147. 6.7506740. 

log 53403 = 4.7275657, 
log 53402 = 4.7275575; 
log 5340234. 6.7275603. 

log 56412 = 4.7513715, 
log 56413 = 4.7513792; 
log 564.123. 2.7513738. 



30. 


Given 


find 
31. 


Given 


and 




32. 


Given 



find 



Ans. 4.9413333. 



PLANE TRIGONOMETRY. 

log 87364 = 4.9413325, 
log 87365 = 4.9413375; 
log .0008736416. 

log 37245 = 4.5710680, 
log 37246 = 45710796; 
log 3.72456. .5710750. 

log 32025 = 4.5054891, 
log 32026 = 4.5055027; 
log 32.025613. 1.5054974. 

log 65931 = 4.8190897, 

log 65932 = 4.8190962; 

log .000006593171. 6.8190943. 

log 25819 = 4.4119394, 

log 25820 = 4.4119562; 

log 2.581926. .4119438. 

log 23454 = 4.3702169, 
log 23453 = 4.3701984; 
log 23453487. 7.3702074. 

log 45740 = 4.6602962, 
log 45741 = 4.6603057; 
find the number whose logarithm is 4.6602987. 



120 




33. 


Given 


find 




34. 


Given 


find 




35. 


Given 


find 




36. 


Given 


find 




37. 


Given 


find 




38. 


Given 


find 




39. 


Given 



- 



40. Given log 43965 = 4.6431071, 

log 43966 = 4.6431170; 
find the number whose logarithm is 4.6431150. .000439658. 



41. Given log 56891 = 4.7550436, 

log 56892 = 4.7550512; 
find the number whose logarithm is .7550480. 



5.689158. 






EXAMPLES. 121 

42. Given log 34572 = 4.5387245, 

log 34573 = 4.5387371; 
find the number whose logarithm is 2.5387359. 

Ans. 345.7291. 

43. Given log 10905 = 4.0376257, 

log 10906 = 4.0376655 ; 
find the number whose logarithm is 3.0376371. 1090.5286. 

44. Given log 25725 = 4.4103554, 

log 25726 = 4.4103723; 
find the number whose logarithm is 7.4103720. 

Ans. .00000025725982. 

In the following six examples the student must take his 
logarithms from the tables. 

45. Kequired the product of 3670.257 and 12.61158, by 
logarithms. Ans. 46287.74. 

46. Required the quotient of ,1234567 by 54.87645, by 
logarithms. Ans. .002249721. 

47. Eequired the cube of .3180236, by logarithms. 

Ans. .03216458. 

48. Eequired the cube root of .3663265, by logarithms. 

Ans. .7155216. 

49. Eequired the eleventh root of 63.742. 1.45894. 

50. Eequired the fifth root of .07. .58752. 



.6737652. 



51. Given 


sin 42° 21'= .6736577, 




sin 42° 22'= .6738727; 


id 


sin 42° 21' 30". 


52. Given 


sin 67° 22'= .9229865, 




sin 67° 23' = .9230984; 


id 


sin 67° 22' 48".5. 



.9230769. 



122 PLANE TRIGONOMETRY. 

53. Given sin 7° 17' = .1267761, 

sin 7° 18' = .1270646 ; 
find sin 7° 17' 25". 

54. Given cos 21° 27' = .9307370, 

cos 21° 28' = .9306306 ; 
find cos 21° 27' 45". 

55. Given cos 34° 12' = .8270806, 

cos 34° 13' = .8269170; 
find cos 34° 12' 19".6. 

56. Given sin 41° 48' = .6665325, 

sin 41° 49' = .6667493; 
find the angle whose sine is .6666666. 

57. Given sin 73° 44' =-.9599684, 

sin 73° 45' = .9600499; 
find the angle whose sine is .96. 

58. Given cos 75° 32' = .2498167, 

cos 75° 31' = .2500984; 
find the angle whose cosine is .25. 

59. Given cos 53° 7' = .6001876, 

cos 53° 8' = .5999549; 
find the angle whose cosine is .6. 

60. Given log sin 45° 16' = 9.8514969, 

log sin 45° 17' = 9.8516220 ; 
find log sin 45° 16' 30". 

61. Given log sin 38° 24' = 9.7931949, 

log sin 38° 25' = 9.7933543 ; 
find log sin 38° 24' 27". 






Ans. .1268963. 



.9306572. 



.8270272. 



41° 48' 37". 



73°44'23".2. 



75° 31' 21". 



53° 7' 48".4. 



9.8515594. 



9.7932666. 



62. 


Given 


log sin 32° 28' = 9.7298197, 
log sin 32° 29' = 9.7300182 ; 




find 




log sin 32° 28' 36". Ans. 


9.7299388. 


63. 


Given 


log sin 17° 1' = 9.4663483. 
log sin 17° 0' = 9.4659353 ; 




find 




log sin 17° 0'12". 


9.4660179. 


64. 


Given 


log sin 26° 24' = 9.6480038. 
log sin 26° 25' = 9.6482582; 




find 




log sin 26° 24' 12". 


9.6480547. 


65. 


Given 


log cos 17° 31' = 9.9793796, 
log cos 17° 32' = 9.9793398 ; 




find 




logcosl7°31'25".2. 


9.9793629. 


66. 


Given 


log tan 21° 17' = 9.5905617, 
log tan 21° 18' = 9.5909351 ; 




find 




log tan 21° 17' 12". 


9.5906364. 


67. 


Given 


log tan 27° 26' = 9.7152419, 
log tan 27° 27' = 9.7155508 ; 




find 




log tan 27° 26' 42". 


9.7154581. 


68. 


Given 


log cot 72° 15' = 9.5052891, 
log cot 72° 16' = 9.5048538 ; 




find 




log cot 72° 15' 35". 


9.5050352. 


69. 


Given 


log cot 36° 18' = 10.1339650, 
log cot 36° 19' = 10.1337003 ; 




find 




log cot 36° 18' 20". 


10.1338768. 


70. 


Given 


log cot 51° 17' = 9.9039733, 
log cot 51° 18' = 9.9037144; 




find 




log cot 51° 17' 32". 


9.9038352. 



124 PLANE TRIGONOMETRY. 

71. Given log sin 16° 19' = 9.4486227, 

log sin 16° 20' = 9.4490540 ; 

find the angle whose log sine is 9.4488105. 

Ana. 16° 19' 2 

72. Given log sin 6° 53' =9.0786310, 

log sin 6° 53' 10"= 9.0788054; 

find the angle whose log sine is 9.0787743. 6° 53' 8". 

73. Given log cos 22° 28' 20"= 9.9657025, 

log cos 22° 28' 10"= 9.9657112 ; 

find the angle whose log cosine is 9.9657056. 22° 28' 16". 

In the following examples the tables are to be used : 

74. Find log tan 55° 37' 53". Ans. 10.1650011. 

75. Find log sin 73° 20' 15". 7. 9.9813707. 

76. Find log cos 55° 11' 12". 9.7565636. 

77. Find log tan 16° 0'27". 9.4577109. 

78. Find log sec 16° 0'27". 10.0171747. 

79. Find the angle whose log cosine is 9.9713383. 

Ans. 20° 35' 16". 

80. Find the angle whose log cosine is 9.9165646. 

Ans. 34° 23' 25". 

81. Find log cos 34° 24' 26". 9.9164762. 

82. Find log cos 37° 19' 47". 9.9004540. 

83. Find log sin 37° 19' 47". 9.7827599. 

84. Findlogtan37°19'47". 9.8823059. 

85. Find log sin 32° 18' 24".6. 9.7279096. 

86. Findlogcos32°18'24".6. 9.9269585. 

87. Findlogtan32°18'24".6. 9.8009511. 



EXAMPLES. 125 

Prove the following by the use of logarithms : 

88. ( 7 - 014 ) 3 - * = .9942207. 
(7.014) 3 + 1 



89 _ V^12 x V.0000307 5 = . 00Q2 3 2432 . 
a/80 -r- a/,0000001 

90 . ^(2002)^x(1001)g =21840300000j 
V 1001 x 2002 



126 PLANE TBIGONOMETBY. 






CHAPTER V. 
SOLUTION OF TKIGONOMETEIO EQUATIONS. 

82. A Trigonometric Equation is an equation in which 
the unknown quantities involve trigonometric functions. 

The solutioyi of a trigonometric equation is the process of 
finding the values of the unknown quantity which satisfy the 
equation. As in Algebra, we may have two or more simul- 
taneous equations, the number of angles involved being 
equal to the number of equations. 



EXAMPLES. 

1 






1. Solve sin0 

2 

This is a trigonometric equation. To solve it we must 

find some angle whose sine is — We know that sin 30°= — 

5 2 2 

Therefore, if 30° be put for 6, the equation is satisfied. 
.•. = 30° is a solution of the equation. 
... = wir + (-l) n ^ (Art. 38) 

2. Solve cos0 + sec<9 = -. 

The usual method of solution is to express all the func- 
tions in terms of one of them. 

Thus, we put for sec 0, and get 

COS0 



cos0 + -^ = -- 

cos0 2 






TRIGONOMETRIC EQUATIONS. 127 

This is an equation in which 0, and therefore cos 0, is 
unknown. We proceed to solve the equation algebraically- 
just as we should if x occupied the place of cos 0, thus : 



COS 2 


--COS0 = -1. 

2 




4 4 




= 2 or - 



The value 2 is inadmissible, for there is no angle whose 
cosine is numerically greater than 1 (Art. 21). 

.-. costf^ 1 - 

2 

But cos 60° = -. 

2 

.\ cos = cos 60°. 
Therefore one value of which satisfies the equation is 60°. 

3. Solve cosec - cot 2 <9 + 1 = 0. 

We have cosec - (cosec 2 - 1) + 1 = . . (Art. 23) 
cosec 2 — cosec = 2. 

.-. cosec0 = -±- 

= 2 or - 1. 
But cosec 30° = 2. 

.*. cosec = cosec 30°. 
Therefore 30° is one value of which satisfies the equation. 

Find a value of which will satisfy the following equa- 
tions : 

4. cos = cos 2 0. Arts. Ibr. 

5. 2 cos = sec ft 45°. 



A 


us. 


60°. 






30°. 






30°. 


0° 


or 


45°. 
60°. 
45°. 




7T 

2' 


2ir 
3' 



128 PLANE TRIGONOMETRY. 

6. 4sin0 — 3cosec0 = O. 

7. 4 cos = 3 sec ft 

8. 3sin0-2cos 2 = O. 

9. V'2sin0 = tan0. 

10. tan = 3 cot 0. 

11. tan + 3 cot = 4. 

12. cos + cos 3(9 + cos 50 = 0. 

13. sin (0 - <£) = i, cos (0 + <£) = 0. = 60°. <£ = 30°. 

83. /SW?;e £7ie equations 

m sin <£ = a (1) 

m cos <£ = & (2) 

where a and b are given, and the values of m and <£ are 
required. 

Dividing (1) by (2), we get 

tan <£ = -, 
b 

which gives two values of <j>, differing by 180% and there- 
fore two values of m also from either of the equations 

a b 

m = — = 

sin cj> cos </> 

The two values of m will be equal numerically with 
opposite signs. 

In practice, m is almost always positive by the conditions 
of the problem. Accordingly, sin cf> has the sign of a, and 
cos <£ the sign of b, and hence <£ must be' taken in the quad- 
rant denoted by these signs. These cases may be considered 
as follows: 

(1) Sin cj> and cos <f> both jiositive. This requires that the 
angle <j> be taken in the first quadrant, because sin cj> and 
cos <j> are both positive in no ether quadrant. 






TRIGONOMETRIC EQUATIONS. 129 

(2) Sin cf> positive and cos <f> negative. This requires that <f> 
be taken in the second quadrant, because only in this quad- 
rant is sin <f> positive and cos <f> negative for f;he same angle. 

(3) Sin <j> and cos <j> both negative. This requires that <£ 
be taken in the third quadrant, because only in this quad- 
rant are sin <j> and cos <£ both negative for the same angle. 

(4) Sin <£ negative and cos <f> positive. This requires that 
<£ be taken in the fourth quadrant, because only in this 
quadrant is sin <f> negative and cos <j> positive for the same 
angle. 

Ex. 1. Solve the equations msin<£ = 332.76, and mcos <j> 
= 290.08, for m and <j>. 

log m sin <£ = 2.52213 
log m cos cj> = 2.46252 






<£ = 48° 55'.2. 



m = 441.45. 



log tan <f> = 0.05961 

log m sin <£ = 2.52213 

log sin </> = 9.87725 

log m = 2.64488 



Ex. 2. Solve m sin <f> = - 72.631, and m cos <£ = 38.412. 

Ans. $ = 117° 52'.3, m = - 82.164. 

84. Solve the equation 

a sin<£ + b cos </> = c . . . . . . (1) 

a, b, and c being given, and cj> required. 

Find in the tables the angle whose tangent is - ; let it 
be/8. a 

Then - = tan /3, and (1) becomes 
a 

a(sin<£ + tan/3 cos </>) = c; 
/sin $ cos /? + cos <£ sin /? 
\ cos/2 

or sin(<£ + /?) = - cos /3 = - sin/? .... (2) 



130 PLANE TRIGONOMETRY. 

There will be two solutions from the two values of <j> + /3 
given in (2) . 

Find from the tables the value of cos /J. Next find from 

the tables the magnitude of the angle a whose sine = - 

a 
cos p, and we get 

sin(<£ + /?) = since, 

.-. d> + p = mr + (-l) n a . . (Art. 38) 

... £ = _£ + „!!. + (_!)»«, 

where n is zero or any positive or negative integer. 

In order that the solution may be possible, it is necessary 

to have - cos ft =, or < 1. 
a 

Note. — This example might have been solved by squaring both sides of the 
equation; but in solving trigonometric equations, it is important, if possible, to 
avoid squaring both sides of the equation. 

Thus, solve cos 9 = k sin 9 (3) 

If we square both sides we get 

cos 2 9 = k 2 sin 2 9 = k 2 (l — cos 2 9) . 

.-. cos 2 0= — ^— ; or cos0 = ± (4) 

k 
Now if a be the least angle whose cosine = — , we get from (4) 

<Sl + &, 

9 = nrr±a (5) 

But (3) may be written cot 9 = k. 

.-. 9 = nn + a (6) 

(6) is the complete solution of the given equation (3) , while (5) is the solution of 
both cos 9 = k sin 0, and also of cos 9 = — k sin 9. Therefore by squaring both mem- 
bers of an equation we obtain solutions which do not belong to the given equation. 

EXAMPLES. 

1. Solve 0.7466898 sin - 1.0498 cos <f> = - 0.431689, 
when <£ < 180°. 

log b = 0.02112-* 
log a = 1.87314 

log tan /S = 014798- 
.-. £ = 125° 25' 20". 

* The minus sign is written thus to denote that it belongs to the natural number 
and does not affect the logarithm. Sometimes the letter n is written instead of the 
minus sign, to denote the same thing. 



TRIGONOMETRIC EQUATIONS. 131 

log sin £=9.91111 
log c = 1.63517 - 
colog&=9.97888- 

log sm(4 + i8)= 9.52516+ 

.-. 4> + p = 19° 34' 40" or 160° 25' 20". 
.-. 4 = - 105° 50' 40" or 35° 0' 0". 

2. Solve - 23.8 sin <f> + 19.3 cos <f> = 17.5(> < 180°). 

4«s. <j> = 4° 12'.7 or - 106° 7'.9. 

3. Solve 2 sin + 2 cos 6 = V2. .4n*. - ^+?itt + (-1) M £- 

4 b 

4. " sin 6> + V3 cos 6 = 1. — \ + mr + (-l)"|. 

5. " sin 6 - cos -0 = 1. ^ + wtt + (- 1) m 1tt. 

4: 

6. " V3 sin 0- cos <9 = V2. ^tt + utt + (- l)"^- 

85. /Sofae ^e equation 

sm(a + x) = msmx (1) 

in which a and m are given. 
Prom (1) we have 

sin (a + x) + sin x _m + 1 
sin (a + #) — sin a? m — 1 



tan f x + - 
V 2 

tan- 



• • (2) 
(Art. 46) 



.\ tan (x + ia) = m tan^ .... (3) 
m — 1 2 

which determines a? + |-«, and therefore x. 

If we introduce an auxiliary angle, the calculation of 
equation (3) is facilitated. 



132 PLANE TRIGONOMETRY. 

Thus, let m = tan<£ ; then we have by [(14) of Art. 61] 

m + l = tan^+1 = cot ( _ ^ 
m — 1 tan <£ — 1 

which in (3) gives 

tan f x + -^ = cot O - 45°) tan^-cc . (4) 

This, with tan <f> = m, 

gives the logarithmic solution. 

The logarithmic solution of the equation 
sin (cc — x) = m sin a; 
is found in the same manner to be 

tan <j> = m, 

and tan (a; — 5 j = cot (<j> + 45°) tan % 

which the student may show. 
Example. — Solve 

sin (106°+ x) = - 1.263 sinx(a; < 180°). 
log tan <£ = logm = log (- 1.263) = 0.10140 -. 
.-. <£ = 128° 22'.3. 
<£--45 = 83°22'.3; log cot (<£ - 45°)= 9.06523 
i a = 53° O'.O, logtan|c* = 10.12289 



log tan /» + -) = 9.18812 



a + ^ = 8°46'.0 or 188° 46'. 
2 

... a = _44°14' or 135° 46 r . 

86. Solve the equation 

tan (a + x) = m tana; (1) 

in which a and m are given. 






TRIGONOMETRIC EQUATIONS. 133 

From (1) we have 

tan ((* + %) + tan x _ m + 1 
t tan (a + x) — tana; m — 1 

sin(cc+2#) 



sin a 



[(21) of Art. 61] 






.\ sin(cg + 2ff) = m ~r" since (2) 

m — 1 

=T^ot (<£ - 45°) sin a . (Art. 85) 
where tan <£ = m. 

Example. — Solve tan (23° 16'+ «) = .296 tana;, 
log tan </> = log m = log (.296) = 1.47129. 
.;. <£ = 16° 29'.3. 
<j> - 45°= - 28° 30'.7 ; log cot(<jf> - 45°) = 10.26502- 
a = 23°16'.l, log sin a = 9.59661 

logsin(a + 2a;) = 9.86163- 
a + 2a; = 226 38'.9 or 313° 21U. 
.-. a;=101 o 41'.5 or 145° 2\6. 

87. Solve the equation 

tan(a + #) tano;= m (1) 

in which a and m are given. 

From (1) we have 
1 + tan (a + x) tan x _ 1 + m 
1 — tan (cc + x) tan x 1 — m 

= 52^ . . (Ex. 4 of Art. 47) 

cos (a + 2x) 

.-. cos(<* + 2x) = - cos a 

1 + m 

= tan (45° - <£) cos a [ (16) of Art. 61] 

phere tan <£ = m. 



134 PLANE TRIGONOMETRY. 

Example. — Solve tan (65° + x) tanx = 1.5196 (a < 180°). 
log tan <j> = logm == 0.18173. 
.-. <£ = 50°39'9"; 
45° - <{> = - 11° 39' 9" ; log tan(45°-<£) = 9.31434- 
a = 65° 0' 0" ; log cos a = 9.62595 

log cos(<* + 2x)= 8.94029- 
a + 2^ = 95° or 265°. 

.-. 2^=30° or 200°. 

.-. a = 15° or 100°. 

88. Solve the equations 

m sin (0 + x) = a . . . . (1) 

m sin (<£ + x) = b (2) 

for m and x, the other four quantities, 6, <jf>, a, b, being 
known. 

Expanding (1) and (2) by (Art. 44), we get 

m sin 6 cos x + m cos 6 sin x = a (3) 

m sin <£ cos x + m cos <£ sin x = b (4) 

Multiplying (3) by sin cf> and (4) by sin 0, and subtracting 
the latter from the former, we have 

m sin x (sin <£ cos 6 — cos <£ sin 6) = a sin $ — b sin 6. 

a sin 6 — 6sin0 /e ,v 

.-. msmx= % \ (5) 

sm(<£-0) v } 

To find the value of mcos#, multiply (3) and (4) by 
cos <£ and cos 6, respectively, and subtract the former from 
the latter. Thus 

m cos #(sin <£ cos 6 — cos <£ sin 6)=b cos 8 — a cos <j>. 

.-. mo,osx = : -£ (o) 

sin(<£-0) v ' 

Having obtained the values of m sin x and m cos x from 
(5) and (6), m and a? can be calculated by Art. 83. 



TRIGONOMETRIC EQUATIONS. 135 



EXAMPLES. 



1. Solve m cos (6 + x) = a, and m sin ( <f> + x ) = 6, for 
m sin x and m cos x. A . 6 cos — a sin 6 

cos(0-4>) 

WC oss = 6siDg + acos *. 

cos (0 — </>) 

2. Solve m cos (0 + a;) = a, and m cos (<j> — x)=b, for 
m sin # and m cos x. A . 6 cos — a cos <f> 

sin (0 + <j>) 

b sin + a sin d> 

m cos a; = ! -*-• 

sin ((9 + <jf>) 

89. Solve the equation 

x cos a + 2/ sin a = m (1) 

a^sina — 2/cos« = n (2) 

for a; and y. 

Multiplying (1) by cos a and (2) by since, and adding, 

we get 

x = m cos a + n sin a. 

To find the value of y, multiply (1) by sin a and (2) by 
cos a, and subtract the latter from the former. Thus 

y = m sin a — n cos a. 

Example. — Solve 

x sin a + y cos a = a, 
x cos a — y sin a=b. 

90. To adapt Formulae to Logarithmic Computation. — 

As calculations are performed principally by means of 
logarithms, and as we are not able by logarithms directly* 
to add and subtract quantities, it becomes necessary to 
know how to transform sums and differences into products 

* Addition and Subtraction Tables are published, by means of which the logarithm 
of the sum or difference of two numbers may be obtained. (See Tafeln der Addi- 
tions, und Subtractions, Logarithmen fur sieben Stellen, von J. Zech, Berlin.) 



136 PLANE TRIGONOMETRY. 

and quotients. An expression in the form of a product or 
quotient is said to be adapted to logarithmic computation. 

An angle, introduced into an expression in order to adapt 
it to logarithmic computation, is called a Subsidiary Angle. 
Such an angle was introduced into each of the Arts. 84, 85, 
86, and 87. 

The following are further examples of the use of sub- 
sidiary angles : 

1. Transform a cos ± b sin into a product, so as to 
adapt it to logarithmic computation. 

Put - = tan<£;* thus 
a 

a cos ±bsmO = a ( cos ± -sin 6 ) 

V a ' J 

= a (cos ± tan <£ sin 0) 
a 



cos <£ 
Similarly, 



cos(0T <£). 



a sin ± b cos = — — sin (0 ± <£). 

cos</> 

3. Transform a ± b into a product, 

a + b = afl + - \ = a(l + tan 2 <£) = asec 2 <£, 

if - = tan 2 <f). 

a 



a — b = all ) = acos 2 <£, 



if ^ = sin 2 <£. 

a 

* The fundamental formulae cos (a? ± y) and sin (a? ± y) (Art. 42) afford examples of 
one term equal to the sum or difference of two terms; hence we may transform an 
expression a cos 9 ± b sin 9 into an equivalent product, by conforming it to the for- 
mulae just mentioned. 

Thus, comparing the identity, m cos <J> cos 9 ± m sin <f> sin 9 = m cos (</> =F 9) or m cos 
(9 + <f>) , with a cos 9 ± b sin 9, we will have a cos 9 ± b sin 9 = m cos (9 t <j>) if we assume 
a = m cos <J> and b = m sin </> ; i.e. (Art. 83) , if tan <£ = - and m = — —■ = ^— - as above. 
See Art. 84. ° C08 * em * 



ADAPTATION TO LOGARITHMIC COMPUTATION. 137 

.\ log (a + 6) = log a + 21ogsec<£; 
and log (a — b) = log a + 2 log cos <£. 

4. Transform 1239.3 sin 6 — 724.6 cos 6 to a product. 

log 6 = log (- 724.6) = 2.86010- 
loga = log (1239.3) = 3.09318 

log tan <j> = 9.76692- 
... $ = _ 30° 18\8. 

loga = 3.09318 
log cos </> = 9.93615 

log-^_ = 3.15703 
cos<£ 

1435.6. 



cos<£ 
.\ 1239.3 sin (9 - 724.6 cos 6 = 1435.6 sin (0 - 30° 18'.8). 

91. Solve the equations 

rcos</> cos0 = a (1) 

rcoscfrsmO = b (2) 

rsin<£ = c (3) 

for r, <£, and 6. 

Dividing (2) by (1), we have 

tan0 = -, 
a 

from which we obtain 0. 
From (1) and (2) we have 

rcos<£ = -^-=:-^- (4) 

cos sin 

from which we obtain rcos <£. 

From (3) and (4) we obtain r and <f> (Art. 83). 



138 



PLANE TRIGONOMETRY. 



EXAMPLES. 



1. Solve 



for r, <£, 0. 



r cos <£ cos 6 = — 53.953, 
rcos<£ sin 6 = 197.207, 
rsin<£ = - 39.062, 



... <£ = -10°49'. 
logr cos <£ = 2.31060 
log cos <£ = 9.99221 

logr = 2.31839 
.% r = 208.16. 



log 6 = 2.29493 
log a = 1.73201 - 

log tan (9 = 0.56292- 

.-. = 1O5°18'.O. 
log sin 6 = 9.98433 

log r cos <£ = 2. 31060 
logr sin = 1.59175- 

log tan <£ = 9.28115- 



92. Trigonometric Elimination. — Several simultaneous 
equations may be given, as in Algebra, by the combination 
of which certain quantities may be eliminated, and a result 
obtained involving the remaining quantities. 

Trigonometric elimination occurs chiefly in the applica- 
tion of Trigonometry to the higher branches of Mathematics, 
as, for example, in Physical Astronomy, Mechanics, Analytic 
Geometry, etc. As no special rules can be given, we illus- 
trate the process by a few examples. 

EXAMPLES. 

1. Eliminate </> from the equations 

x = a cos <£, y = b sin </S. 
From the given equations we have 



- = COS d>, ^ : 

a b 



■ sin <t>. 



which in 
gives 



cos 2 <£ + sin 2 <£ = 1, 



x 2 



y; 



9 " 7.9 — - 1 - 



a' 



TRIGONOMETRIC ELIMINATIONS. 139 

2. Eliminate <f> from the equations 
a cos <j> + b sin <f> = c, 
b cos </> + c sin cf> = a. 

Solving these equations for sin <£ and cos <f>, we have 

. , 6c — a 2 

sm<£ = - , 

o z — ac 

„„„ , c 2 — ab 

cos <£ = ; 

ac — er 

which in cos 2 <£ + sin 2 cj> = 1, 

yives (be - a 2 ) 2 + (c 2 - ab) 2 = (ac - 6 2 ) 2 . 

3. Eliminate <£ from the equations 

y cos cf> — x sin cj> = a cos 2 <£, 
2/sin<£ + #cos<£ = 2asin2<£. 

Solve for x and 2/, then add and subtract, and we get 
x + y = a(sin<£ + cos<£) (1 + sin2<£), 
x — y = a(sin <£ — cos <£) (1 — sin2 </>). 

... (x + 2 /) 2 = a 2 (l + sin2^>) 3 , 
(x-y) 2 = a 2 (l-$m2<t>)\ 

... (aj + y)£ + (aj-y) ? = 2a* 

4. Eliminate a and (3 from the equations 
a = since cos /? sin 6 + cos a cos 6 . . . (1) 
b = sin a cos ft cos 6 — cos a sin . . . (2) 
c = sin a sin/8 sin0 (3) 

Squaring (1) and (2), and adding, we get 

a 2 + b 2 = sin 2 a cos 2 /3 + cos 2 a (4) 

c 2 
. n = sin 2 « sin 2 /? , . (5) 



140 PLANE TRIGONOMETRY. 

Adding (4) and (5), we have 

a 2 + b 2 + -^-=l. 
snrO 

5. Eliminate <j> from the equations 

a sin <jf> + b cos <£ = c, 

a cos <j> — b sin <£ = d. Ans. a 2 + b 2 *= c 2 + d 2 . 

6. Eliminate 6 from the equations 

m = cosec 6 -— sin 0, 

n = sec — cos 0. mW (m ¥ + n*) = 1. 

7. Eliminate 6 and <£ from the equations 

sin 6 + sin <£ = a, 
cos0 + cos<£ = by 
cos (<9 - <£ ) = c. a 2 + 6 2 - 2 c = 2. 

8. Eliminate x and ?/ from the equations 

tan x + tan y = a, 

cot a? + cot y = b, 

x + y = c. cot c = • 

a 6 

9. Eliminate <£ from the equations 

x = cos 2 <£ + cos <£, 
y = sin 2 <£ + sin <f>. 

Ans. 2x = (x 2 + y 2 ) 2 -3(x 2 + y 2 ). 

EXAMPLES. 

Solve the following equations : 

1. tan + cot 6 = 2. <4ns. 45°. 

2. 2 sin 2 <9 + V2 cos = 2. 90°, or 45°. 

3. 3tan 2 <9-4sin 2 <9 = l. 45°. 

4. 2sin 2 6>+V2sin0 = 2. 45°. 



EXAMPLES. 141 

5. cos 2 0-V3cos + f = O. Ans. 30°. 

6. sin 5 = 16 sin 5 6. mr, or m? ± -• 

6 

7. sin9 — sin 0= sin 40. Jwtt, or £wtt ± — • 

io 

8. 2 sin = tan 0. nir, ox 2mr± -• 

' 3 

9. 6cot 2 6»-4cos 2 = l. mr±-' 

3 

10. tan + tan (<9 - 45°) = 2. »ir±-. 

11. cos0 + V3sin0 = V2. 2rnr±-- 

12. tan ((9 + 45°) = 1 + sin 2 0. mr- -. or mr. 

13. (cot0-tan0) 2 (2 + V3) = 4(2-V3). £n7r± — . 

14. cosec cot = 2 V3. 2mr±- 

6 

15. cosec + cot = V3. 2mr + \ir. 



16. sin- = cosec — cot 0. 2mr. 

17. sin 5(9 cos 30 = sin 9(9 cos 70. -^mr + (-1)-. 

18. sin 2 (9 + cos 2 2 (9 = f . ?itt ± ?-, or n«r ± T \ v. 

19. V3sin0-cos0 = V2. riTr + - + (- l) w -. 

6 4 

20. tan + cot = 4. nTr + ^-Tr. 

21. sin(0 + <£)=^cos(0-</>) = ^. • = j+-^ 

22. Solve m sin <£ = 1.29743, and ra cos <£ = 6.0024. 

-4ri5. O<180°). 



142 PLANE TRIGONOMETRY. 

23. Solve m sin <£= -0.3076258, and m cos <£= 0.4278735. 
(m positive.) Arts. <j> = 324° 17' 6".6, m == 0.52698. 

24. Solve m sin c/> = 0.08219, and m cos <£ = 0.1288. 

25. Solve m sin <£ = 194.683, and m cos <£ = 8460.7. 

26. If asin0 + 6cos0 = c, and a cos 6 + 6 sin 5 = c sin0 
cos 0, show that sin 2 0(c 2 — a 2 — 6 2 ) = 2 a&. 

Solve the following equations : 

27. V2sin0 + V2cos<9 = V3. Ans. -j + nw + (- l) n ^- 

28. 2 sin a? + 5 cos x = 2. Sug. [2.5 = tan 68° 12']. 

Ans. x = -68° 12' + n 180° + ( - l) w (21° 48'). 

29. 3cosa-8sina = 3. Sug. [2.6 = tan 69° 26' 30"]. 

Ans. x = - 69° 26' 30" + 2 n 180° ± (69° 26' 30"). 

30. 4 sin x - 15 cos x = 4. Sug. [3.75 = tan 75° 4']. 

-4ns. ® = 75° 4' + n 180° + ( - l) w (14° 56') . 

31. cos (a + x) = sin (a + x) + V2 cos /?. 

Ans. x = —a — ^ + 2n7r±(S. 

32. cos + cos 3 6 + cos 5 <9 = 0. 

J^is. !(2w + 1)tt, or i(3n±l)7r. 

33. sin5<9 = sin3<9 + sin0 = 3-4sin 2 <9. 

Ans. nw±-, or -i-(2n + 1)tt. 
o 



34. 2 sin 2 3 6 + sin 2 6 = 2. 

35. a(cos20-l) + 2&(cos0 + l) = O. 



Ans. i(2» + l>, or ^ + (_1)-X. 



J.ns. (2n + l)7r, or cos - " 1 -. 

6d 






EXAMPLES. 143 

36. Solve 

m sin(0 + x) = a cos /3, and m cos (0 — x) = a sin /?, 
for m sin x and m cos x. (Art. 67.) 

^Ins. m sin x = \PjJl — L 

cos 2 ' 

a sin (B — 6) 
m cos x = ^- £-• 

cos 2 

37. Solve racos(0 + <£) = 3.79, and m cos (0 — <£) = 2.06, 
for m and 0, when <£ = 31°27'.4. (Art. 67.) 

\[ 38. Solve r cos <f> cos (9 = 1.271, 

r cos <£ sin = — 0.981, 
r sin <£ = 0.890, 
for r, & 0. (Art. 70.) 

39. Solve r cos <£ cos = — 2, 

r cos <£ sin 6 = + 3, 

r sin cf> = — 4, 
for r, <£, 0. 

40. Solve r sin <£ sin = 19.765, 

r sin <£ cos = — 7.192, 

r cos <£ = 12.124, 
for r, <£, 0. 

41. Solve cos(2x + 3y) = ± cos (3 a; + 2?/) = -|V3. 

42. Solve cos3 4-cos50+V2(cos0 + sin0)cos0 = O. 
^tis. 40±0 = 2n7r±f7r, or |(27i + l)7r. 

43. Solve cos3 + sin3 = cos0 + sin0. 
Arts, sin = 0, or tan = — 1 ± V2. 

44. Solve 3sin0 + cos0 = 2#, sin0 + 2cos0 = #. 

.4**. = 71° 34', z = ivT0, 



144 PLANE TBIGONOMETBY. 

45. Solve 1.268 sin <£ = 0.948 + m sin (25° 27'.2), 

1.268 cos <£ = 0.281 + m cos (25° 27'.2). 

Ans. <£ = 60°53'.8, m = 0.372. 

46. Transform x* + y* + z A — 2 y 2 z 2 — 2 z 2 v? — 2 arfy 2 into a 
product. Ans. --(x+y+z)(y+z—x){z-\-x—y)(x+y—z). 

47. Eliminate from the equations 
m sin 2 = n sin 0, p cos 20 = q cos 0. 

-4ns. m 2 +p 2 = n 2 + g 2 . 

48. Eliminate and <£ from the equations 
x = a cos m cos m <j>, y = b cos m sin m </>, z = c sin™ 0. 

w + w + w 

49. Eliminate from the equations 
a sin + 6 cos 6= 7i, a cos — 6 sin 6 = Jc. 

Ans. a 2 + b 2 = h 2 + 7c 2 . 

50. Eliminate from the equations 
a tan + 6 sec = c, a f cot + 6' cosec = c f . 

.^ws. (a'b + c6') 2 + (a6 r + c'6) 2 == (ce' - aa') 2 . 

51. Eliminate from the equations 
x = 2 a cos cos 2 — a cos 0, 
y = 2a cos sin 2 — a sin 0. ^4ns. x 2 + y 2 = a 2 . 

52. Eliminate from the equations 
x = a cos + 6 cos 2 8, and ?/ = a sin + 6 sin 2 0. 

.4ws. a 2 [(a; + &)■ + y 2 ] = [x 2 + y 2 - 6 2 ] 2 . 

53. Eliminate a and /? from the equations 
6 + c cos a = u cos (a — 6), 
6 + ccos/J = ^ cos (/? — 0), a — P = 2cf>; 

and show that 

w 2 — 2 wc cos + c 2 = 6 2 sec 2 <£. 



EXAMPLES IN ELIMINATION. 145 

54. Eliminate and <f> from the equations 

x cos + y sin = a, b sin (0 + <£) = a sin <£, 
a; cos (0 + 2 <£) — ?/ sin (0 + 2<j>) = a. 

Ans.x 2 + y 2 = a 2 + c ^* 
u b 2 

55. Eliminate from the equations 

x __ sec 2 — cos 2 
a sec 2 + cos 2 0' 

?5 = sec 2 + cos 2 0. ^ + ^ = 1. 

y a 2 b 2 

56. Eliminate from the equations 

(a + 6) tan (0 — <£) = (a — b) tan (0 + <£), 

acos2<£ + 6 cos 20 = c. ^Lns. 6 2 =c 2 +a 2 — 2 ac cos 2 <£. 

57. Eliminate from the equations 

cos 2 , sin 2 1 



x sin — y cos = -y/x 2 + 2/ 2 , — - — f 



2 a^ + 2/ 2 

^n«. ^ + ^=1. 
a 2 6 2 

58. Eliminate and <f> from the equations 

a 2 cos 2 — b 2 cos 2 <f> = c?, acos0 + 6cos<£ = r, 
a tan = 5 tan <j>. 

4r 2 a 2 



Ans. a 2 



■.■M^V*} 






(V + c 2 ) 2 

59. Eliminate <f> from the equations 
n sin — m cos = 2 m sin <£, 

n sin 26 — m cos 2 <£ = n. 

Ans. (nsin0 + mcos0) 2 ==2m(m + n). 

60. Eliminate a from the equations 
x tan (a — (3) = y tan (<* + /?), 

(# — 2/) cos 2 a -f (a; + ?/) cos 2 /? = 2. 

^Ins. z 2 + 4 a;?/ = 2z (x -f- 2/) cos 2/3. 



146 



PLANE TRIGONOMETRY* 



CHAPTER VI. 



KELATIONS BETWEEN THE SIDES OF A TKIANGLE 
AND THE FUNCTIONS OF ITS ANGLES. 

93. Formulae. — In this chapter we shall deduce formula) 
which express certain relations between the sides of a tri- 
angle and the functions of its angles. These relations will 
be applied in the next chapter to the solution of triangles* 
One oi' the principal objects of Trigonometry, as its name 
implies (Art. 1), is to establish certain relations between 
the sides and angles oi' triangles, so that when some of 
these are known the rest may be determined. 

RIGHT TRIANGLES. 

94. Let ABC be a triangle, right-angled at C, 
the angles of the triangle by the let- 
ters A, B, C, and the lengths of the 
sides respectively opposite these an- 
gles, by the letters a, b } C* Then we 
have (Art. 14) the following relations : a b ^C 

a as csin A = ccos B = b tan A = frcotB . . (1) 

b = csinB = ceos A = a tan B = a cot A . . (2) 

# c = bwe A = (( see I> = ftcosec B = a cosec A . (<*>) 

which may be expressed in the following general theorems: 

* The student must remember that a, l>, o, are numbers expressing the lengths of 

the sides In terms Of BOme unit Of length, such an s foot OF a mile. The unit may be 
whatever we please, but must be the same for all the aides. 




OBLIQUE TBI ANGLES. 



147 



I. In a right triangle each side is equal to the product of the 
hypotenuse into the sine of the opposite angle or the cosine of 
the adjacent angle. 

J I. In a right triangle each side is equal to the product of 
the other .side into the tangent of the angle adjacent to that 
other side, or the cotangent of the angle adjacent to itself 

III. In a right triangle the hypotenuse is equal to tJie 
product of a -side into the secant of Us adjacent angle, or the 
cosecant of its opposite angle. 

EXAMPLES. 

In a right triangle ABC, in which C is a right angle, 
prove the following : 

1. tan B = cot A + cos C. 
3. cos2A + cos2B = 0. 

5. cosec 2 d = - H 

2b 2a 



7. tan2A; 



2ab 



b 2 - a 2 



2. 


sin2A = 


= sin2B. 


4. 


Bin2A = 


_ 2 aft 

o 2 


6. 


cos 2 A = 


_ b 2 - a 2 ^ 

c 2 


8. 


sin 3 A - 


3a&*-a? 



OBLIQUE TRIANGLES. 

95. Law of Sines. — In any triangle the sides are pro- 
portional to the sines of the opposite angles. 

Let ABC be any triangle. Draw 
CD perpendicular to AD. 

We have, then, in both figures 

CD = a sin B = h sin A. (Art. 94) 

.-. a sin B = b sin A. 

a b 

sin A sin B 

Similarly, by drawing a perpen- 
dicular from A or B to the opposite 

side, we may prove that 




148 



PLANE TRIGONOMETRY. 



or 



sin B sin C 
a b 



, and 



sin C sin A 



sin A sin B sin C 
a : b : c = sin A : sin B : sin C. 



96. Law of Cosines. — In any triangle the square of any 
side is equal to the sum of the squares of the other two sides 
minus twice the product of these sides and the cosine of the 
included angle. 

In an acute-angled triangle (see 
first figure) we have (Geom., Book 
III., Prop. 26) 

BC 2 = AC 2 + AB 2 - 2 AB x AD, 
or a 2 = 6 2 + c 2 -2c.AD. 
But AD = b cos A. 

,\ a 2 = b 2 + c 2 — 2 be cos A. 

In an obtuse-angled triangle (see 
second figure) we have (Geom., 
Book III., Prop. 27) 

BC 2 = AC 2 + AB 2 + 2 AB x AD, 

or a 2 = 6 2 + c 2 + 2c.AD. 

But AD = b cos CAD = — b cos A. 

,\ a 2 = 6 2 + c 2 — 2 6c cos A. 

Similarly, b 2 = c 2 + a 2 — 2 ca cos B, 

c 2 = a 2 + & 2 -2a&cosC. 

Note. — When one equation in the solution of triangles has been obtained, the 
other two may generally be obtained by advancing the letters so that a becomes b, 
b becomes c, and c becomes a', the order is abc, bca, cab. It is obvious that the 
formulae thus obtained are true, since the naming of the sides makes no difference, 
provided the right order is maintained. 







OBLIQUE TRIANGLES. 149 

97. Law of Tangents. — In any triangle the sum of any 

two sides is to their difference as the tangent of half the sum 
of the opposite angles is to the tangent of half their difference. 

By Art. 95, a:b = sin A : sin B. 

By composition and division, 

a + b _ sin A + sin B 
a — ■ b sin A — sin B 

= tan *(A + B) b (13) f A 61 (1) 

Similarly. M^ = tanj(B + C) (2) 

b-c tan£(B-C) V ; 

c + a _ tan j-(C + A) ( ~\ 

c-a"~tan|(C — A) ^ 

Since tan£(A + B) = tan(90° - £C) = cot ^C, 

the result in (1) may be written 

a + b _ cot^C /n 

a-&""tan£(A-B) ^ ) 

and similar expressions for (2) and (3). 

98. To show that in any triangle c = a cos B + b cos A. 

In an acute-angled triangle (first figure of Art. 96) we 
nave c = DB + DA 

= a cos B + b cos A. 

In an obtuse-angled triangle (second figure of Art. 96) 
we have C = DB-DA 






= a cos B — b cos CAD. 
.-. c = acosB + 6 cos A. 
Similarly, b = c cos A + a cos C, 

a = 6cosC + ccosB. 



150 PLANE TRIGONOMETRY. 



EXAMPLES. 

1. In the triangle ABC prove (1) 

a + b:c = cos \ (A — B) : sin \ C, 
and (2) a — b : c = sin i (A — B) : cos \ C. 

2. If AD bisects the angle A of the triangle ABC, prove 

BD:DC = sinC:sinB. 

3. If AD' bisects the external vertical angle A, prove 

BD':CD' = sinC:sinB. 

, tt 1 2cosiAcosi(B-C) 

4. Hence prove _ = ^— ^ h 

and also -i- = 2 8in|Asin^(C-B) a 

D'C asinB 

99. To express the Sine, the Cosine, and the Tangent of 
Half an Angle of a Triangle in Terms of the Sides. 

I. By Art. 96 we have 

cos A = V + ° 2 ~" a2 = 1 - 2 sin 2 A . . . . (Art. 49) 

2bc 2 v ; 



2 26c 

= qg-(ft- C )» 

2 be 

__ (a + b — c)(a — b + c) 
~ 2bc 

Let a + b + c = 2s; 

then a + b — c = 2(s—c), and a — b + c = 2(s — b)„ 

. 2 A 2(s-c)2(s-6) 
.\ 2sm 2 — = — * y — ^ 1~. 

2 2bc 



sin 



A a / (—&)(—£ . . . . (1) 

2 \ 6c w 



OBLIQUE TRIANGLES. 151 

Similarly, sm| = .^iZ^ES (2) 



V (S ~H S " 6) (3) 



sin C = jis-aHs-b) 
2 \ a& 



II. cosA = 2cos 2 ^-l (Art. 49) 

Li 

9 A ^ , 6 2 + c 2 — a 2 

.\ 2 cos 2 — =1H — 

2 26c 

^ jb + cY-a 2 

2 be 

_ (a + b + c)(b + c — a) 

" 26c 

= 2s-2(s-q) 

26c 



A 

COS-—: 

2 \ 6c 



; 



=\^ m 

Similarly, cos— - = A r^~ ' (5) 

2 \ ac 

C ls(s — c) //lx 

COS 2 = V^5 (6) 

III. Dividing (1) by (4), we get 

^l=^W^ (7) 

Similarly, tan| = V«^ (*> 

taqg=J( & - ffl )( s - 6 ) (9) 

2 \ s(s-c) V ' 

Since any angle of a triangle is < 180°, the half angle is 

< 90° ; therefore the positive sign must be given to the 
adicals which occur in this article. 



152 PLANE TRIGONOMETRY. 

100. To express the Sine of an Angle in Terms of the 
Sides. 

sinA = 2sin|:cos^ (Art. 49) 

z z 

= 2c f (s ~^0 ? ~ c) " ./ *(«-«) 
\ 6c V 6c 

(Art. 99) 

O 

.•. sinA = — Vs(s — a)(s — b)(s — c). 
be 



Similarly, sin B = — Vs (s — a) (s — b) (s — c), 
ac 

2 



sin C = — Vs (s — a) (s — 6) (s — c) . 
ab 



Cor. sin A = — V26 2 c 2 +2c 2 a 2 + 2 a 2 6 2 - a 4 - 6 4 - c 4 

26c 

and similar expressions for sin B, sin C. 



EXAMPLES. 

In any triangle ABC prove the following statements : 

1. a (b cos C — c cos B) = b 2 — c 2 . 

2. (6 + c) cos A + (c + a) cos B + (a + b) cos C = a + b + a 

o sin A + 2 sin B __ sin C 
a + 2b c 

A sin 2 A — m sin 2 B sin 2 C 

4. ~ — = — . 

a 2 — mb 2 <? 

5. a cos A + b cos B — c cos C = 2c cos A cos B. 

6 cos A cosB cosC « 

sin B sin C sin C sin A sin A sin B ~ 

7. a sin (B - C) + b sin (C - A) + c sin (A - B) = 0. 

8. tan£Atan£B = ^i. 

s 

9. tan£A-^tan£B = (s-&)-5-($--c). 






AREA OF A TRIANGLE. 



153 



101. Expressions for the Area 
of a Triangle. 

(1) Given tivo sides and their 
included angle. 

Let S denote the area of the tri- 
angle ABC. Then by Geometry, 

28 = ex CD. 



But in either figure, by Art. 



94, 






CD = b sin A. 



,\ S = ^6csinA. 




Similarly, S = ^ ac sin B, 

S = \ ab sin C. 

(2) Given one side and the angles. 



Since 



which is 



a : b = sin A : sin B , (Art. 95) 

, a sin B 
sm A 

S = ^ ab sin C, gives 

a 2 sin B sin C 



Similarly, 



S = 
S = 



2 sin A 
b 2 sin A sin C c 2 sin A sin B 



2sinB 



2sinC 



(3) Given the three sides. 



sin A = iL V«(* - a) (s -b)(s- c) (Art. 100) 
oc 









Substituting in 



we get 



S = \ be sin A, 

S = Vs(s — a)(s — b)(s — c). 



154 



PLANE TBIGONOMETBY. 



102. Inscribed Circle. — To find the radius of the inscribed 
circle of a triangle. q 

Let ABC be a triangle, the 
centre of the inscribed circle, and 
r its radius. Draw radii to the 
points of contact D, E, F; and join 
OA, OB, OC. Then A c D 

S = area of ABC 

= A AOB + A BOC + A COA 

= ±rc + ±ra + ± rb 




a + b + c 
= r — ! ! — = rs 



-u 



(s — a)(s — b) (s — c) 



(Art. 99) 
(Art. 101) 



103. Circumscribed Circle. — To find the radius of the 
circumscribed circle of a triangle in ^^—^^JZ 

terms of the sides of the triangle. 

Let be the centre of the circle A/ 
described about the triangle ABC, 
and E its radius. 

Through draw the diameter CD 
and join BD. 

Then Z BDC = Z BAC = Z A. 

.\ BC = 2EsinA, or a = 2EsinA 



E = 



2 sin A 2 sin B 2 sin C 



But 



. A 2S 

sin A = — 

be 

-r> abc 
•'• R= 4S 




RADII OF THE ESCRIBED CIRCLES. 



155 



104. Escribed Circle. — To find 
the radii of the escribed circles of a 
Mangle. 

A circle, which touches one side 
of a triangle and the other two 
sides produced, is called an escribed 
circle of the triangle. 

Let be the centre of the 
escribed circle which touches the 
side BC and the other sides pro- 
duced, at the points D, E, and F, 
respectively, and let the radius of 
this circle be r x . 

We then have from the figure 

AABC = AAOB + AAOC-ABOC. 

. Q _ CT l i ^ T l aT l 

= |r 1 (6 + c-a) = r 1 (s-a) . 
S 




ri =- 



a 



(Art. 99) 
. . (1) 



Similarly it may be proved that if r 2 , r 3 are the radii of 
the circles touching AC and AB respectively, 

S S 



n = 



6' 



r* = - 



105. To find the Distance be- 
tween the Centres of the Inscribed 
and Circumscribed Circles * of a 
Triangle. 

Let I and be the incentre 
and circumcentre, respectively, of 
the triangle ABC, IA and IC 
bisect the angles BAC and BCA ; 

* Often called the incentre and circumcen- 
tre of a triangle. 




156 PLANE TRIGONOMETRY. 

therefore the arc BD is equal to the arc DC, and DOH 
bisects BC at right angles. 

Draw IM perpendicular to AC. Then 

Z DIC = A±S1 = BCD + BCI = DCL 

.-. DI = DC = 2Esin-. 

2 

Also, AI = IM cosec— = r cosec — 

* 2 2 

.-. DI.AI = 2Rr = EI-IF; 
that is, (E + 01) (E - 01) = 2 Br. 

.-. OI 2 = E 2 -2Er. 



EXAMPLES. 

1. The sides of a triangle are 18, 24, 30 ; find the radii 
of its inscribed, escribed, and circumscribed circles. 

Ana. 6, 12, 18, 36, 15. 

2. Prove that the area of the triangle ABC is 

1 c 2 

2 cot A + cot B' 

3. Find the area of the triangle ABC when 

(1) a = 4, 6 = 10 ft., C = 30°. Arts. 10 sq. ft. 

(2) 6=5, c = 20 inches, A = 60°. 43.3 sq. in. 

(3) a = 13, b = 14, c = 15 chains. 84 sq. chains. 

A T> 1111 

4. Prove - = — -| 1 



5. Prove r = g-g- in ^ Bsin ^ C 



cos^- A 



EXAMPLES. 



157 



6. Prove that the area of the triangle ABC is represented 
by each of the three expressions : 

2 R 2 sin A sin B sin C, 

rs, and 

Br (sin A + sin B + sin C). 

7. If A = 60°, a = V3, 6 = V2, prove that the area 
= i(3+V3). 

8. Prove E (sin A + sin B + sin C) = s. 

9. Prove that the bisectors of the angles A, B, C, of a 
triangle are, respectively, equal to 



A B 

2 be cos — 2 ea cos — 2ab cos 



C 



b + e 



e + a 



a + b 






106. To find the Area of a Cyclic * 
Quadrilateral. 

Let ABCD be the quadrilateral, and 
a, b, c, and d its sides. Join BD. - 
Then, area of figure == S 

= ±ad sin A + \ be sin C 
= |(ad + 6c)sinA ... (1) 
NowinAABD, BD 2 = a 2 + d 2 - 2ad cos A, 
and in A CBD, BD 2 = b 2 + c 2 - 2 be cos C 

= & 2 + c 8 -2&ccosA. 

«™ a a 2 -6 2 -c 2 + d 2 

. \ cos A = — ■ — • 

2 (ad + be) 




sin A 



=v-[- 



_ y _ <? + cpr 
2 {ad + be) 



_ V(2ad + 25c) 2 - (a 2 - 6 2 - c 2 + d 2 ) 1 
2(ad + bc) 

* See Geometry, Art. 251. 






i + ftc) 



SuUstitut - o* hare 

S x ^ «)(*-») (*-■ 

N more important ahapter ate 

I up as t, - 

* = ?>*- i-,vs\ 

a — 6 taa£(A — B) 



I feiA-^*=*M»=2> (Art 

v \ \ 






\ 



l€ 

gg - - \ «... — H—(* —- 



8. AroaofA=x i*-*)(«-t) 



EZAMPL1 159 

9. Area of A = Ua + h + c ) = ,-. s - .... (A it. 102) 
0. r = J ( * - °)('-&)(*-T) 



11- B = f| (Art. 108) 



EXAMPLES. 

Iii a right triangle ABC, in which C is the right angle, 
prove; the following: 

i. cos2r. = s " r ' A - sl,|2|: . 

sin- A -f sin- 1 J 

2. sm-' ,! = r -« 
2 2c 

4. cos*— = 



2 2c 
5. Bin(A-B) + cos2A = a 

0. = tan 

a + b 2 

7. siii(A-I!) + sin(2A + C) = 0. 

8. tanlA=-^-. 

& + c 

9. (sin A - sin B)*+ (cos A + cos B) 1 = 2. 



10 ^ j a + & i /a — 6 _ 2sinA 



ra + 6 Vcos2B 
In any triangle ABC, prove the following statements: 

11. (a + 6)8in? = ceo8— ~— • 
■2 2 



160 
12. 

13. 
14. 

15. 

16. 

17. 
18. 
19. 

20. 
21. 
22. 
23. 
24. 
25. 
26. 



PLANE TRIGONOMETRY. 

t% \ A • B-C 

(o — c) cos — = asm 

v ' 2 2 

a(6 2 +c 2 )cosA+6(c 2 +a 2 )cosB+c(a 2 +6 2 )cosC=3a6c. 

a — b _ cos B — cos A 
c 1 + cos C 

6 + c cos B + cos C 



a 



1 — cos A 



/t — = — =5 — : — 7i 6 2 sin C + c 2 sin B 

V be sin B sm C = — ■ — • 

b + c 

a+b+c = (6 + c)cos A + (c + a)cos B + (a + 6)cos C. 
6+c— a = (6 + c)cosA — (c — a)cosB + (a — 6)cosC. 
a cos (A+ B + C) — b cos (B + A) - c cos (A + C) = 0. 
cos A cos B cos C _ a 2 + b 2 + c 2 



a 
tanA = 



b c 

asinC 
b — a cos C 



2abc 



COS" 5 h C COS — = 5. 

2 2 

, B, C 6 + c-a 
tan— tan— = — ! 

2 2 6 + c + ce 

A B 

tan— (b + c — a) = tan— (c + a — 6). 
2 ^ 

c 2 = (a + 6) 2 sin 2 - + (a - 6) 2 cos 2 -. 



c(cos A + cos B) = 2 (a -f b) sin 
c(cos A — cos B) = 2(6 — a) cos 



-a 

2 
C 



27. 

28. tan B + tanC = (a 2 + 6 2 - c 2 ) -«- (a 2 - 6 2 + c 2 ). 






EXAMPLES. 161 

29. a 2 + b 2 + <? = 2(ab cos C + be cos A + ea cos B). 

30. cos 2 ---cos 2 | = (s-a)-r-6(s-6). 

- 31. 6 sin 2 — hesin 2 — = s — a. 

32. If jp is the length of the perpendicular from A on 

BC, sinA = ^. 

be 



36 -a 



33. If A = 3B, then sin B = i^ 

34. If V6c sin B sin C = 52sin B + ° 2sin C , then B = C. 

b + e 

ok B C A 

35. a cos— cos — cosec — = s. 

2 2 2 

36. If cos A = f , and cos B = |f , then cos C = — £$. 

37. If sin 2 B + sin 2 C = sin 2 A, then A = 90°. 

38. If D is the middle point of BC, prove that 

4AD 2 =26 2 + 2c 2 -a 2 . 

39. If a = 2b, and A = 3B, prove that C = 60°. 

40. If D, E, F, are the middle points of the sides, BC, CA, 
AB, prove 

4(AjJ 2 + BE* + CF 2 ) = 3(a 2 + b 2 + c 2 ). 

41. If a, b, c, the sides of a triangle, are in arithmetic 
progression, prove 

4. A. C 1 

tan— tan— = — 

2 2 3 

ao t-p tan A — tan B e — b .<, , A /> AO 

4J. If = ■ prove that A = 60 . 

tan A + tan B e 

43. If cos B = ™^., prove that B = C. 

2smC 



162 PLANE TRIGONOMETRY. 

44. If a 2 = b 2 - be + c 2 , prove that A = 60°. 

45. If the sides of a triangle are a, 6, and Va 2 + ab + b 2 y 
prove that its greatest angle is 120°. 

46. Prove that the vertical angle of any triangle is 
divided by the median which bisects the base, into seg- 
ments whose sines are inversely proportional to the adja- 
cent sides. 

47. If AD be the median that bisects BC, prove (1) 

, (&2 _ c 2) tan ADB = 2 bc sin A, 

and (2) cot BAD + cot DAC = 4 cot A + cot B + cot C. 

48. Find the area of the triangle ABC when a = 625, 
b = 505, c = 904 yards. Arts. 151872 sq. yards. 

49. Find the radii of the inscribed and each of the 
escribed circles of the triangle ABC when a = 13, 6 = 14, 
c = 15. Ans. 4; 10.5; 12; 14. 

50. Prove the area S = \ a 2 sin B sin C cosec A. 



51. " " " "=vnwv 

w u u u u 2abc /ABC 

52. " " " " = f cos — cos — cos 



a + b + c\ 2 2 2j 

53. Prove that the lengths of the sides of the pedal tri- 
angle, that is, the triangle formed by joining the feet of the 
perpendiculars, are a cos A, b cos B, c cos C, respectively. 

54. Prove that the angles of the pedal triangle are, 
respectively, ir — 2 A, w — 2B, tt — 2C. 

55. Prove rvr 2 r* = r 3 cot 2 — cot 2 — cot 2 — 

123 2 2 2 

A B P 

56. Prove r t cos — = a cos — cos — 

1 2 2 2 

57. Prove that the area of the incircle : area of the trt 

, ,A ,B ,C 

angle : : tt : cot — cot — cot — 

8 2 2 2 









EXAMPLES. 163 

Prove the following statements : 

58. If a, b, c, are in A. P., then ac = 6rE. 

59. If the altitude of an isosceles triangle is equal to the 
base, E is five-eighths of the base. 

60. 6c = 4E 2 (cosA + cosBcosC). 

61. If C is a right angle, 2r + 2E = a + 6. 

62. r 2 r 3 + r z i\ + r x i\ = s 2 . 

63. I + ! 4 1 1 



be ca ab 2rE 

Q 

64. ri+r 2 = ccot — 

1 2 2 

e - A . B • C 

6o. r cos — = a sin — sin — 

2 2 2 



66. If _p l5 p 2 , p 3 be the distances to the sides from the 



eircumcentre, then 



2l _i_ — -j- c a ^ c 



Pi Pi Ps ^PiPiPs 
67. The radius E of the circumcircle 



: i^ 



abc 



2 \ sin A sin B sin C 



68. S = -sin2B + -sin2A. 

4 4 

69. -l_+_I_ + _l,_l = i5. 

s — a 5 — 6 s — c 5 S 

70. after = 4E(s- a) (s- 6) (s-c). 

71. The distances between the centres of the inscribed 
and escribed circles of the triangle ABC are 4Esin— , 

4Esin- 4Esin-. 

2 ? 2 

72. If A is a right angle, r 2 + r 3 = a. 



164 PLANE TRIGONOMETRY. 

73. In an equilateral triangle 3 E = 6 r = 2 r t . 

74. If r, r 1? r 2 , r s denote the radii of the inscribed and 
escribed circles of a triangle, 

tan 2 - = — i- 

2 r 2 r 3 

75. The sides of a triangle are in arithmetic progression, 
and its area is to that of an equilateral triangle of the same 
perimeter as 3 is to 5. Find the ratio of the sides and the 
value of the largest angle. Ans. As 7, 5, 3 ; 120°. 

76. If an equilateral triangle be described with its 
angular points on the sides of a given right isosceles 
triangle, and one side parallel to the hypotenuse, its area 
will be 2 a 2 sin 2 15° sin 60°, where a is a side of the given 
triangle. 

77. If h be the difference between the sides containing 
the right angle of a right triangle, and S its area, the diam- 
eter of the circumscribing circle = -y/h 2 + 4S. 

78. Three circles touch one another externally : prove 
that the square of the area of the triangle formed by join- 
ing their centres is equal to the product of the sum and 
product of their radii. 

79. On the sides of any triangle equilateral triangles are 
described externally, and their centres are joined: prove 
that the triangle thus formed is equilateral. 

80. If 1? 2 , 3 are the centres of the escribed circles of 
a triangle, then the area of the triangle 1 2 3 = area of 

a , b , c 



triangle ABC 



1 + 



;) 



b + c — a a + c — b a + b — c_ 

81. If the centres of the three escribed circles of a tri- 
angle are joined, then the area of the triangle thus formed 

is — -, where r is the radius of the inscribed circle of the 
2r 

original triangle. 



SOLUTION OF TRIANGLES. 165 



CHAPTER VII. 

SOLUTION OP TRIANGLES. 

107. Triangles. — In every triangle there are six elements, 
the three sides and the three angles. When any three ele- 
ments are given, one at least of the three being a side, the 
other three can be calculated. The process of determining 
the unknown elements from the known is called the solution 
of triangles. 

Note. — If the three angles only of a triangle are given, it is impossible to deter- 
mine the sides, for there is an infinite number of triangles that are equiangular to 
one another. 

Triangles are divided in Trigonometry into right and 
oblique. We shall commence with right triangles, and shall 
suppose C the right angle. 

RIGHT TRIANGLES. 

108. There are Four Cases of Eight Triangles. 

I. Given one side and the hypotenuse. 

II. Given an acute angle and the hypotenuse. 

III. Given one side and an acute angle. 

IV. Given the two sides. 

Let ABC be a triangle, right-angled 
at C, and let a, 6, and c, as before, be 
the sides opposite the angles A, B, 
and C, respectively. 

The formulae for the solution of right triangles are (1), 
(2), (3) of Art. 94. 




166 



PLANE TRIGONOMETRY. 



109. Case I. — Given a side and the hypotenuse, as a and 
c; to find A, B, b. 

We have 






sin A = — 



. \ log sin A = log a — log c, 
from which A is determined ; then B = 90° — A. 
Lastly, b = c cos A. 

.-. log b = log c + log cos A. 
Thus A, B, and b are determined. 
Ex. 1. Given a = 536, c = 941 ; find A, B, b. 



We have 



Solution by Natural Functions. 
.569607. 



• A a 536 
sin A = - = 



c 941 
From a table of natural sines we find that 

A = 34° 43' 22". .-. B = 55° 16' 38". 
Lastly, b = c cos A = 941 x .821918 

= 773.425. 

Logarithmic Solution. 



log sin A = log a — log c. 

log a = 2.7291648 
log c = 2.9735896 

log sin A* = 9.7555752 

.-. A = 34° 43' 22". 

... B = 55° 16' 38". 



log b = log c + log cos A. 

log c = 2.9735896 
log cos A = 9.9148283 

log b = 2.8884179 

.-. 6 = 773.424. 



Our two methods of calculation give results which do not 
quite agree. The discrepancies arise from the defects of 
the tables. 



* Ten is added so as to get the tabular logarithms (Art. 76). 



RIGHT TBIANGLES. 167 

The process of solution by natural sines, cosines, etc., can 
be used to advantage only in cases in which the measures 
of the sides are small numbers. 

We might have determined b thus : 



6 = V(c — a) (c + a); 

or thus : b = a tan B. 

Note. — It is generally better to compute all the required parts from the given 
ones, so that if an error is made in determining one part, that error will not affect the 
computation of the other parts. 

To test the accuracy of the work, compute the same parts by different formulae. 

Ex. 2. Given a = 21, c = 29 ; find A, B, b. 

Ans. A = 46° 23' 50", B = 43° 36' 10", 6 = 20. 

Note. — In these examples the student must find the necessary logarithms from 
the tables. 

110. Case II. — Given an acute angle and the hypotenuse, 
as A and c; to find B, a, b. 

We have B = 90°-A. . 

Also a = c sin A, and b = c cos A. 

.*. log a = log c + log sin A; 
and log b = log c + log cos A. 

Thus B, a, and b are determined. 

Ex. 1. Given A = 54° 28', c = 125 ; find B, a, b. 
B = 90° - A = 35° 32'. 

Solution by Natural Functions. 

We have a = c sin A, and b = c cos A. 

Using a table of natural sines, we have 

a = 125 x .813778 = 101.722, 
b = 125 x .581177 = 72.647. 



168 



PLANE TRIGONOMETRY. 



Logarithmic Solution, 



log a = log c + log sin A. 

log c =2.0969100 
log sin A = 9.9105057 

loga* = 2.0074157 

.-. a = 101.722. 



log b = log c + log cos A. 
log c = 2.0969100 



log cos A = 9.7643080 



: 1.8612180 
: 72.647. 



log b * : 
.\ 6: 



Ex. 2. Given A = 37° 10', c = 8762 ; find a and b. 

Ans. 5293.4; 6982.3. 

111. Case III. — Given a side and an acute angle, as A 
and a; to find B, b, c. 

We have B = 90° - A. 

a -, a 



Also 



b = 



and c = 



and 



tan A sin A 

log b = log a — log tan A, 
log c = log a — log sin A. 



Ex. 1. Given A = 32° 15' 24' 



: 5472.5; find B, b, c. 



Solution. 
B = 90° - A = 57° 44 f 36". 



log b = log a — log tan A. 

log a = 3.7381858 
log tan A = 9.8001090 

log b = 3.9380768 

.-. 6 = 8671.152. 



log c = log a — log sin A. 

log a = 3.7381858 
log sin A = 9.7273076 

log c = 4.0108782 

.-. c= 10253.64. 



Ex. 2. Given A = 34° 18', a = 237.6 ; find B, b, c. 

Ans. B = 55°42'; 6 = 348.31; c = 421.63. 



* Ten is rejected because the tabular logarithmic functions are too large by ten 
(Art 76). 



RIGHT TRIANGLES. 



169 






112. Case IV. — Given the tivo sides, as a and b; to find 

A, B, c. 



We have 
Also 



tan A = -: then B = 90° - A. 
b J 

c = a cosec A = 



sin A 

.-. log tan A = log a — log b, 
and log c = log a — log sin A. 

- Ex. Given a = 2266.35, b = 5439.24 ; find A, B, c. 

Solution. 



log tan A = log a — log b. 

log a = 3.3553270 
los: b = 3.7355382 



log c === log a — log sin A. 

log a = 3.3553270 
log sin A = 9.5850266 

log tan A = 9.6197888 log c = 3.7703004 

A = 22° 37' 12". 

B = 67° 22' 48". 

Note. — In this example we might have found c by means of the formula 
c = Va 2 + b' 2 ; but we would have had to go through the process of squaring the 
values of a and b. If these values are simple numbers, it is often easier to find c in 
this way; but this value of c is not adapted to logarithms. A. formula which consists 
entirely of factors is always preferred to one which consists of terms, when any of 
those terms contain any power of the quantities involved. 

113. When a Side and the Hypotenuse are nearly Equal. 

— When a side and the hypotenuse are given, as a and c 
in Case L, and are nearly equal in value, the angle A is very 
near 90°, and cannot be determined with much accuracy 
from the tables, because the sines of angles near 90° differ 
very little from one another (Art. 81). It is therefore 
desirable, in this case, to find B first, by either of the 
following formulae : 

B /l— cosB 
sm-=-\/- (Art. 50) 



-4 



2 



c — a 
~2c~ 



(1) 



170 PLANE TRIGONOMETRY. 



B /^ - CQsB (Art. 50) 

2 \l + cosB v ' 



tan 

+ cosB 



4 



c ~ a (2) 

lc + a 

Then b = c cos A (3) 

or =V(c + a)(c — a) (4) 

Ex. 1. Given a = 4602.21059, c = 4602.836 ; find B. 

c-a = 0.62541, log (c-a) = 1.7961648 

2 c = 9205.672, log 2 c = 3.9640555 

2)5.8321093 
log sin | = 7.9160547 

B = 56' 40".36. .-. ? = 28'20".18. 

Note. — The characteristic 5 is increased numerically to 6 to make it divisible 
by 2 (see Note 4 of Art. 66). Ten is then added to the characteristic 3, making it 7, 
so as to agree with the Tables (Art. 76). 

There is a slight error in the above value of B on account 
of the irregular differences of the log sines for angles near 
0° (Art. 81). A more accurate value may be found by the 
principle that the sines of small angles are approximately 
proportional to the angles (Art. 130) . 

EXAMPLES. 

The following right triangles must be solved by log- 
arithms. 

1. Given a = 60, c = 100 ; find A, B ? b. 

Ans. A = 36° 52'; B = 53°8'; 6 = 80. 

2. Given a = 137.66, c = 240 ; find A, B, b. 

Ans. A = 35° ; B = 55° ; 6 = 196.59. 



BIGHT TRIANGLES. 171 

3. Given a = 147, c = 184 ; find A, B, 6. 

Ans. A = 53°r35"; B = 36° 58' 25" ; 6 = 110.67. 

4. Given a = 100, c = 200 ; find A, B, 6. 

Arts. A = 30°; B = 60°; b = 100 V3. 

5. Given A = 40°, c = 100 ; find B, a, 6. 

Ans. B = 50°; a = 64.279; 6 = 76.604. 

6. Given A = 30°, c = 150 ; find B, a, b. 

Ans. B = 60°; a = 75; b = 75 V& 

7. Given A = 32°, c = 1760 ; find B, a, 6. 

^4w«. B = 58°; a = 932.66 ; 6 = 1492.57. 

8. Given A = 35° 16' 25", c = 672.3412; find B, a, b. 

Ans. B = 54° 43' 35"; a = 388.26; 6 = 548.9. 

9. Given A = 75°, a = 80 ; find B, 6, c. 

Ans. B = 15°; 6 = 80(2-V3); c = 80(V6 -V2). 

10. Given A = 36°, a = 520 ; find B, 6, c. 

Ans. B = 54° ; 6 = 715.72 ; c = 884.68. 

11. Given A = 34° 15', a = 843.2 ; find B, 6, c. 

Ans. B = 55°45'; c = 1498.2. 

12. Given A = 67° 37' 15", 6 = 254.73 ; find B, a, c. 

Aiis. B = 22° 22' 45" ; a = 618.66 ; c = 669.05. 

13. Given a = 75, 6 = 75 ; find A, B, c. 

Ans. A = 45° = B ; c = 75 V2. 

14. Given a = 21, 6 = 20 ; find A, B, c. 

^7is. A = 46° 23' 50"; c = 29. 

15. Given a = 300.43, 6 = 500 ; find A, B, c. 

^.ns. A = 31°; B = 59° ; c = 583.31. 

16. Given a = 4845, 6 = 4742; find A, B, c. 

Ans. A = 45° 36' 56". 



172 PLANE TRIGONOMETRY. 

OBLIQUE TRIANGLES. 

114. There are Four Cases of Oblique Triangles. 

I. Given a side and tivo angles. 
II. Given two sides and the angle opposite one of them. 

III. Given two sides and the included angle. 

IV. Given the three sides. 

The formulae for the solution of oblique triangles will be 
taken from Chap. VI. Special attention must be given to 
the following three, proved in Arts. 95, 96, 97. 

a b c 



(1) The Sine-rule, 



(2) The Cosine-rule, cos A = 



sin A sin B sin C 



2 be 



(3) The Tangent-rule, tan = cot — • 

2 a + b 2 

115. Case I. Given a side and two angles, as a, B, C ; find 
A, b, c. 

(1) A = 180° - (B + C). .-. A is found. 
b a , asinB 



(2) 



sinB sin A sin A 



/ox c a asinC 

(o) =• • .*. c = • 

sinC sin A sin A 

These determine b and c. 

Ex. 1. Given a = 7012.6, B = 38° 12' 48", C = 60°; find 
A, 6, c. 

Solution. 

A = 180° - (B + C) = 81° 47' 12". 



loga = 3.8458791 

log sin B = 9.7914038 

colog* sin A = 0.0044775 

log b = 3.6417604 

.-. 6 = 4382.89. 



loga = 3.8458791 

log sin C = 9.9375306 

colog sin A = 0.0044775 

log c = 3.7878872 

.-. c= 6136.03 



* See Art. i 



OBLIQUE TBI ANGLES. 173 

Ex. 2. Given a=1000, B = 45°, C m 127° 19 f ; find A, b, c. 
Ans. A = 7° 41'; 6 = 5288.8; c = 5948.5. 

116. Case II. Given tivo sides and the angle opposite one 
of them, as a, b, A ; find B, C, c. 

(1) sinB = — ; thus B is found. 

a 

(2) C = 180° - (A + B); thus C is found. 

(3) c = ^—^5 — ; thus c is found. 

sin A 

This is usually known as the ambiguous case, as shown in 
geometry (B. II., Prop. 31). The ambiguity is found in 
the equation 

^ frsinA 
sin B = 



Since the angle is determined by its sine, it admits of 
two values, which are supplements of each other (Art. 29). 
Therefore, either value of B may be taken, unless excluded 
by the conditions of the problem. 

I. If a < b sin A, sin B > 1, which is impossible ; and 
therefore there is no triangle with the given parts. 

II. If a — b sin A, sin B = 1, and B = 90° ; therefore there 
is one triangle — a right triangle — with the given parts. 

III. If a > b sin A, and < b, sinB < 1 ; hence there are 
two values of B, one being the supplement of the other, 
i.e., one acute, the other obtuse, and both are admissible ; 
therefore there are two triangles with the given parts. 

IV. If a > &, then A > B, and since A is given, B must 
be acute ; thus there is only one triangle with the given 
parts. 



174 



PLANE TRIGONOMETRY. 



These four cases may be illustrated geometrically. 

Draw A, the given angle. Make AC = b ; draw the per- 
pendicular CD, which = b sin 
A. With centre C and radius 
a, describe a circle. 

I. If a<b sin A, the circle 
will not meet AX, and there- 
fore no triangle can be formed 
with the given parts. 

II. If a = b sin A, the cir- 
cle touches AX in B'; there- 
fore there is one triangle, 
right-angled at B. 

III. If a > b sin A, and 
< b, the circle cuts AX in 
two points B and B', on the _\ 
same side of A ; thus there B x 
are two triangles ABC and AB'C, each having the given 
parts, the angles ABC, AB'C being supplementary. 

IV. If a > b, the circle cuts AX on opposite sides of A, 
and only the triangle ABC has the given parts, because the 
angle B'AC of the triangle AB'C is not the given angle A, 
but its supplement. 

These results may be stated as follows : 




a < b sin A, 

a = b sin A, 

a > b sin A and < b, 

a>b, 



no solution. 

one solution (right triangle). 

two solutions. 

one solution. 



These results may be obtained algebraically thus : 



We have 



a 2 = b 2 + c 2 -2bccosA 



(Art. 96) 



b cos A ± V a 2 ■ 



b 2 sin 2 A, 



OBLIQUE TRIANGLES, 175 

giving two roots, real and unequal, equal or imaginary, 
according as a >, =, or < b sin A. 

A discussion of these two values of c gives the same 
results as are found in the above four cases. We leave 
the discussion as an exercise for the student. 

Note. — When two sides and the angle opposite the greater are given, there can 
be no ambiguity, for the angle opposite the less must be acute. 

When the given angle is a right angle or obtuse, the other two angles are both 
acute, and there can be no ambiguity. 

In the solution of triangles there can be no ambiguity, except when an angle is 
determined by the sine or cosecant, and in no case whatever when the triangle has 
a right angle. 

Ex. 1. Given a = 7, 6 = 8, A = 27° 47' 45"; find B, C, c. 

m 



Solution. 



log 6 = 0.9030900 

log sin A = 9.6686860 

colog a = 9.1549020 

log sin B = 9.7266780 



log a = 0.8450980 

log sin C = 9.9375306 

colog sin A = 0.3313140 

log c = 1.1139426 



. •. B = 32° 12' 15", or 147° 47' 45". . \ c = 13. 

.-. C = 120°, or 4° 24' 30". 

Taking the second value of C as follows : 

log a = 0.8450980 

log sin C = 8.8857232 

colog sin A = 0.3313140 

log c = 0.0621352 

.-. c = 1.1538. 

Thus, there are two solutions. See Case III. 

Ex.2. Given a = 31.239, 5 = 49.5053, A = 32° 18'; find 
B, C, c. Ans. B = 56° 56' 56".3, or 123° 3' 3".7 ; 

C = 90°45' 3".7, or 24°38'56".3: 
c = 58.456, or 24.382. 



176 



PLANE TRIGONOMETRY. 



117. Case III. — Given two sides and the included angle, 
as a, b, C; find A, B, c. 



(i) 



, A — B a — b , C / \ a. -in a\ 

tan ^- = ^ cot 2 • -A^m 



Hence is known, and — — — = 90° — — • 

2 ; 2 2 



. A and B are found. 

a sinC 



(2) 



c = 



or 



b sinC 



sin A ' sinB 

and thus c is found and the triangle solved. 

In simple cases the third side c may be, found directly by 
the formula 



c = Va 2 + & 2 -2a&cosC . . . (Art. 96) 

or the formula may be adapted to logarithmic calculation 
by the use of a subsidiary angle (Art. 90). 

Ex.1. Given a = 234.7, 6 = 185.4, C = 84°36'; find A, 
B, c. 

Solution. 

log(a-6)= 1.6928469 



a = 234.7 

b = 185.4 



a- 6= 49.3 
a + b = 420.1 

.-. ^ = 42° 18'. 

2 



colog(a+ b)= 7.3766473 

log cot - = 10.0409920 

s 2 



log tan 



A-B 



2 
A-B 



= 9.1104862 



= 7° 20' 56". 



A + B 



. 47° 42'. 



A = 55° 2' 56", 
B = 40°21' 4", 
C = 84° 36', 
c = 285.0745. 



log b = 2.2681097 

log sin C = 9.9980683 

colog sin B = 0.1887804 

log c = 2.4549584 






OBLIQUE TRIANGLES. 177 



Ex. 2. Given a = .062387, b = .023475, C = 110° 32' ; find 
A, B, c. 

Ans. A = 52° 10' 33" ; B = 17° 17' 27" ; c = .0739635. 



118. Case IV. Given the three sides, as a, b, c;foid A, B, C. 



The solution in this case may be performed by the for- 
• mulee of Art. 99. By means of these formulae we may 
compute two of the angles, and find the third by subtract- 
ing their sum from 180°. But in practice it is better to 
compute the three angles independently, and check the 
accuracy of the work by taking their sum. 

If only one angle is to be found, the formulae for the sines 
or cosines may be used. If all the angles are to be found, 
the tangent formulae are the most convenient, because then 
we require only the logarithms of the same four quantities, 
s, s — a, s — b, s — c, to find all the angles ; whereas the sine 
and cosine formulae require in addition the logs of a, 6, c. 

The tangent formulae (Art. 99) may be reduced as follows : 



tan A^ Ks—bH^A 

2 \ s(s- a) 



rbv 



(s — a)(s — b)(s — c) 



s — a m s 



... tanA = _JL_ (Art. 102) 



Similarly, tan — = > 

J ' 2 s-b 



tan? = - r 



2 s-c 

Note. — The quantity r is the radius of the inscribed circle (Art. 102). 



178 PLANS TBlGONOMirrnr. 

Ex. 1. G ivon a = 13, b = 14, c = 15 ; liml A, B, C. 

Solution. 

a = 13 log (a - a) = .9030900 

6s« 14 log (*-/;)= .8460980 

c = L6 log(s-c) = .7781513 



2 S = 42 00 1°S 8 = 8.6777807 

8l _21. log » J = 1.2041200 

r-a = 8 logr = .6020600. 

a -6 = 7, .*. log tan I = 9.6989700. 

* ~ c = (> ' ••• log ban ! : = 9.7569620. 



•> 



A = 53° 7' 4S".3S ; 
B = 69°29'28".18; 
C = 67°22'48".44. 



log tan ~ = 9.8239087. 



Without the use of logarithms, the angles may be found 
by the eosine formula* (Art. 96), These may sometimes 
be used with advantage, when the given lengths of a, b, c 
each OOntain less than three digits. 

Ex. 2. Find the greatest angle m the triangle whose 
sides are 13, 11, L5. 

Let a — If), 6 bb 14, c sb L3. Then A is the greatest angle. 

Then eo , A .y + o'-q' = 14' + iy-15? 

•2 be 2 X II \13 

^ = .384618 = oos 67 8 23', nearly 
(by the table of natural Bines). 

.-. the greatest angle is (>7° 23'. 






EXAMPLES. 179 

EXAMPLES. 

1. Given a = 254, B = 16°, C = 64°; find 6 = 71.0919. 

2. Given c = 338.65, A = 53° 24', B = 66°27'; find 
a = 313.46. 

3. Given c = 38, A = 48°, B = 54°; find a = 28.87, 
6 = 31.43. 

4. Given a = 7012.5, B = 38° 12' 48", C = 60°; find 
b and c. ^ MS . & _ 4382.82 ; c = 6135.94. 

5. Given a = 528, 6 = 252, A = 124° 34'; find B and G 

Ans. B = 23°8'33"; C = 32° 17' 27". 

6. Given a = 170.6, 6 = 140.5, B = 40°; find A and G 

Ans. A = 51° 18' 21", or 128° 41' 39"; 
C = 88° 41' 39", or 11° 18' 21". 

7. Given a = 97, 6 = 119, A = 50°; find B and C. 

Ans. B = 70° 0' 56", or 109° 59' 4"; 
C = 59° 59' 4", or 20° 0' 56". 

8. Given a = 7, 6 = 8, A = 27° 47' 45" ; find B, C, c. 

Ans. B = 32° 12' 15", or 147° 47' 45" ; 
C = 120°, or 4° 24' 30" ; 

c = 13, or 1.15385. 

9. Given 6 = 55, c = 45, A = 6° ; find B and G 

Ans. B = 149° 20' 31"; C = 24° 39' 29". 

10. Given 6 = 131, c = 72, A = 40° ; find B and C. 
Ans. B = 108° 36' 30" ; C = 31° 23' 30". 

11. Given a = 35, 6 = 21, C = 50° ; find A and B. 
Ans. A = 93° 11' 49" ; B = 36° 48' 11". 

12. Given a = 601, 6=289, C = 100°19'6"; find A and B. 
Ans. A = 56° 8' 42" ; B = 23° 32' 12". 



180 PLANE TRIGONOMETRY. 

13. Given a = 222, 6 = 318, c = 406 ; find A = 32° 57' 8". 

14. Given a = 275.35, 6=189.28, c=301.47 ; find A, B, C. 
Ans. A = 63° 30' 57" ; B = 37° 58' 20" ; C = 78° 30' 43". 

15. Given a = 5238, 6 = 5662, c = 9384 ; find A and B. 

Ans. A m 29° 17' 16" ; B = 31° 55' 31". 

16. Given a = 317, b = 533, c = 510 ; find A, B, C. 
Ans. A = 35° 18'©"; B = 76° 18' 52" ; C = 68°23'8". 

119. Area of a Triangle (Art. 101). 

EXAMPLES. 
Find the area : 

1. Given a = 116.082, 6 = 100, C = 118° 15' 41". 

Ans. 5112.25. 

2. Given a = 8, 6 = 5, C = 60°. 17.3205. 

3. Given 6 = 21.5, c = 30.456, A = 41° 22'. 2m372. 

4. Given a = 72.3, A = 52° 35', B = 63° 17'. 2644.94. 

5. Given 6 = 100, A = 76° 38' 13", C = 40° 5'. 3506.815. 

6. Given a = 31.325, B = 13° 57' 2", A = 53° 11' 18". 

Ans. 135.3545. 

7. Given a = .582, 6 = .601, c = .427. .117655. 

8. Given « = 408, 6 = 41, c = 401. 8160. 

9. Given a = .9, 6 = 1.2, c = 1.5. .54. 

10. Given a = 21, 6 = 20, c = 29. 210. 

11. Given a = 24, 6 = 30, c = 18. 216. 

12. Given a = 63.89, 6 = 138.24, c = 121.15. 3869.2. 



HEIGHT OF AN OBJECT. 181 



MEASUREMENT OF HEIGHTS AND DISTANCES. 

120. Definitions. — One of the most important applica- 
tions of Trigonometry is the determination of the heights 
and distances of objects which cannot be actually measured. 

The actual measurement, with scientific accuracy, of a 
line of any considerable length, is a very long and difficult 
operation. But the accurate; measurement of an angle, with 
proper instruments, can be made with comparative ease and 
rapidity. 

By the aid of the Solution of Triangles we can determine : 

(1) The distance between points which are inaccessible 

(2) The magnitude of angles which cannot be practically 
observed. 

(3) The relative heights of distant and inaccessible 
points. 

A vertical line is the line assumed by a plummet when 
freely suspended by a cord, and allowed to come to rest. 

A vertical plane is any plane containing a vertical line. 

A horizontal plane is a plane perpendicular to a vertical 
line. 

A vertical angle is one lying in a vertical plane. 

A horizontal angle is one lying in a horizontal plane. 

An angle of elevation is a vertical angle having one side 
horizontal and the other ascending. 

An angle of depression is a vertical angle having one side 
horizontal and the other descending. 

By distance is meant the horizontal distance, unless other- 
wise; named. 

By height is meant the vertical height above or below the 
horizontal plane of the observer. 

For a description of the requisite instruments, and the 
method of using them, the student is referred to books on 
practical surveying.* 

* See Johnson's Surveying, Gillespie's Surveying, Clarke's Geodesy, Gore's 
Geodesy, etc. 



182 PLANE TRIGONOMETRY. 

121. To find the Height of an Object standing on a 
Horizontal Plane, the Base of the Object 

being Accessible. 

Let BC be a vertical object, such as y^ 

a church spire or a tower. y^ 

From the base C measure a horizon- £ 
tal line CA. 

At the point A measure the angle of elevation CAB. 

We can then determine the height of the object BC ; for 

BC = ACtanCAB. 

EXAMPLES. 

1. If AC = 100 feet and CAB = 60°, find BC. 

Arts. 173.2 feet. 

2. If AC = 125 feet and CAB = 52° 34', find BC. 

Ans. 163.3 feet. 

3. AC, the breadth of a river, is 100 feet. At the point 
A, on one bank, the angle of elevation of B, the top of a tree 
on the other bank directly opposite, is 25° 37' ; find the 
height of the tree. Ans. 47.9 feet. 

122. To find the Height and Distance of an Inaccessible 
Object on a Horizontal Plane. 






,A° 



V 



<$ 



Let CD be the object, whose base D 
is inaccessible ; and let it be required 
to find the height CD, and its horizon- 
tal distance from A, the nearest acces- B 
sible point. 

(1) At A in the horizontal line BAD observe the Z DAC 
= a ; measure AB = a, and at B observe the Z DBC = (3. 

Then CA = a sin ^ (Art. 95) 

sin (a - P) V ' 



HEIGHT OF AN OBJECT. 



183 



and 



.-. CD = CAsin«= CTsin '* sin f, 

sm (a — p) 

» t^ a-, a sin B cos a 
AD = AC cos a = ^ - 

sm(a — /?) 



(2) JF/ien £7*e line BA cannot be measured directly 
the object. 

At A observe the vertical Z CAD 
= a, and the horizontal Z DAB = f3 ; 
measure AB = a, and at B observe 
the Z DBA = y. 

Then AD = asin ? . 

sin(/? + y) 

.-. CD = AD tan a 

_ a sin y tan a 
~sin(/? + y) ' 




EXAMPLES. 

1. A river 300 feet wide runs at the foot of a tower, 
which subtends an angle of 22° 30' at the edge of the remote 
bank ; find the height of the tower. Arts. 124.26 feet. 

2. At 360 feet from the foot of a steeple the elevation is 
half what it is at 135 feet ; find its height. Ans. 180 feet. 

3. A person standing on the bank of a river observes the 
angle subtended by a tree on the opposite bank to be 60°, 
and when he retires 40 feet from the river's bank he finds 
the angle to be 30° ; find the height of the tree and the 
breadth of the river. Ans. 20 V3; 20. 

ta 



4. What is the height of a hill whose angle of elevation, 

I taken at the bottom, was 46°, and 100 yards farther off, on 
a level with the bottom, the angle was 31° ? 

Ans. 143.14 yards. 



184 PLANE TRIGONOMETRY. 

123. To find the Height of an Inaccessible Object situated 
above a Horizontal Plane, and its 

C 



Height above the Plane. 

Let CD be the object, and let 
A and B be two points in the y^ / ,/ 

horizontal plane, and in the same y* &\ 

vertical plane with CD. s^^ y® x x 



A 



/ * 



At A, in the horizontal line B a A ^ 

B AE, observe the A C AE = a, 

and DAE = y; measure AB == a, and at B observe the 
ZCBE = ft. 

Then CE = asm " slD/? (Art. 122) 

sin (<*-/?) v ' 

Also, AE = ^cos<*sinft (Art. 122) 

sin (<*-/?) v ' 

T)"p _ a cos a sin ft tan y 
sin (a — ft) 

-,-p, a sin 3 c • . > 

.-. CD = ^ — sin« — cos a tan v( 

sm(<*-ft)* l% 

- a s * n P s * n ( a ~ y) 

cos y sin (a — ft) 
EXAMPLES. 

1. A man 6 feet high stands at a distance of 4 feet 9 
inches from a lamp-post, and it is observed that his shadow 
is 19 feet long : find the height of the lamp. Arts. 7-J feet. 

2. A flagstaff, 25 feet high, stands on the top of a cliff, 
and from a point on the seashore the angles of elevation 
of the highest and lowest points of the flagstaff are observed 
to be 47° 12' and 45° 13' respectively : find the height of the 
cliff. Ana. 348 feet. 

3. A castle standing on the top of a cliff is observed 
from two stations at sea, which are in line with it; their 




HEIGHT OF AN OBJECT. 185 

distance is a quarter of a mile : the elevation of the top 
of the castle, seen from the remote station, is 16° 28'; the 
elevations of the top and bottom, seen from the near 
station, are 52° 24' and 48° 38' respectively: (1) what is 
its height, and (2) what its elevation above the sea ? 

Ans. (1) 60.82 feet ; (2) 445.23 feet. 

124. To find the Distance of an Object on a Horizontal 
Plane, from Observations made at Two Points, in the Same 
Vertical Line, above the Plane. 

Let the points of observation A 
and B be in the same vertical line, 
and at a given distance from each 
other ; let C be the point observed, 
whose horizontal distance CD and 
vertical distance AD are required. 

Measure the angles of depression, 6BC, aAC, equal to a 
and f3 respectively, and denote AB by a. 

Then BD = CDtan<x, AD = CDtan/3. 

i\ a = CD (tan a — tan /?). 

p-p. __ a cos a cos /? 

sin (a — (3) ' 

, A -^ a cos a sin 3 

and AD = — — p • 

sin (a -r- (3) 

EXAMPLES. 

1. From the top of a house, and from a window 30 feet 
below the top, the angles of depression of an object on the 
ground are 15° 40' and 10° : find (1) the horizontal distance 
of the object, and (2) the height of the house. 

Aiis. (1) 288.1 feet; (2) 80.8 feet. 

2. From the top and bottom of a castle, which is 68 feet 
ligh, the depressions of a ship at sea are observed to be 
16° 28' and 14° : find its distance. Ans. 570.2 yards. 




186 PLANE TRIGONOMETRY. 

125. To find the Distance between Two Inaccessible Objects 
on a Horizontal Plane. 

Let C and D be the two inac- 
cessible objects. 

Measure a base line AB, from 
whose extremities C and D are 
visible. At A observe the angles 
CAD, DAB ; and at B observe 
the angles CBA and CBP. 

Then, in the triangle ABC, we know two angles and the 
side AB. .-. AC may be found. In the triangle ABD we 
know two angles and the side AB. .-. AD may be found. 

Lastly, in the triangle ACD, AC and AD have been deter- 
mined, and the included angle CAD has been measured ; 
and thus CD can be found. 

EXAMPLES. 

1. Let AB = 1000 yards, the angles BAC, BAD = 76° 30' 
and 44° 10', respectively ; and the angles ABD, ABC = 
81° 12' and 46° 5', respectively : find the distance between 
C and D. Ans. 669.8 yards. 

2. A and B are two trees on one side of a river ; at two 
stations P and Q on the other side observations are taken, 
and it is found that the angles APB, BPQ, AQP are each 
equal to 30°, and that the angle AQB is equal to 60°. If 
PQ = a, show that 

AB = -V21. 
6 

126. The Dip of the Horizon. — Since the surface of the 
earth is spherical, it is obvious that an object on it will be 
visible only for a certain distance depending on its height; 
and, conversely, that at a certain height above the ground 
the visible horizon will be limited. 



THE DIP OF THE HORIZOX. 



187 



Let be the centre of the earth, P a point above the sur- 
face, PD a tangent to the surface at q_ 
D. Then D is a point on the terres- 
trial horizon ; and CPD, which is the 
angle of depression of the most dis- 
tant point on the horizon seen from 
P, is called the dip of the horizon at P. 
The angle DOP is equal to it. 

Denote the angle CPD by 0, the 
height AP by h, and the radius OD 
by r. Then 




ft = OP- 


-OA = 


r sec 6 — r 






r (1 — cos 6) 
cos 6 




.\ r — 


ft cos 6 
1 — cos 6 


PD = r 


tan0 — 


h sin 8 
1 — cosQ 



=z1i cot ^ . . (Art. 48 ? Ex. 8) 

Also PD 2 = PA x PB = h (h + 2 r) . . . (Geom.) 

Since, in all cases which can occur in practice, h is very 
small compared with 2 r, we have approximately 

TT) 2 = 2hr. 

Let n = the number # of miles in PD, h = the feet in PA, 
and r = 4000 miles nearly. Then 



h = 



PD _ (5280 n) 2 



2r 8000x5280 



5280 n 2 5.28 2 2 2 
— — ?r = -n. 



8000 



I* It will be noticed that n is a number merely, and that the result will be in feet, 
since the miles have been reduced to feet. 



188 



PLANE TRIGONOMETRY. 



That is, the height at which objects can be seen varies as 
the square of the distance. 

Thus, if n = 1 mile, we have 

h = | feet = 8 inches ; 
if n = 2 miles, h = f • 2 2 = § feet, etc., etc. 

Thus it appears that an object less than 8 inches above 
the surface of still water will be invisible to an eye on the 
surface at the distance of a mile. 

Example. From a balloon, at an elevation of 4 miles, the 
dip of the sea-horizon is observed to be 2° 33' 40" : find (1) 
the diameter of the earth, and (2) the distance of the 
horizon from the balloon. 

Ans. (1) 8001.24 miles; (2) 178.944 miles. 

127. Problem of Pothenot or of Snellius. — To determine 
a point in the plane of a given triangle, at which the sides of 
the triangle subtend given angles. 

Let ABC be the given triangle, and 
P the required point. Join P with 
A, B, C. 

Let the given angles APCJ BPC be 
denoted by a, /}, and the unknown 
angles PAC, PBC by x, y respectively ; 
then a and /} are known ; and when x 
and y are found, the position of P can 
be determined, for the distances PA 
and PB can be found by solving the 
triangles PAC, PBC. 

We have x + y = 2tt — a — /} — C . . . . 

Also bsmx = asm 1==Fa 

sin a sin f$ 
Assume an auxiliary angle <£ such that 

, , a sin a 
tan $ = , ; 
b sin/} 

then the value of <£ can be found from the tables. 




(i) 



EXAMPLES. 189 

Thus, 5H1E = tan <f>. 

smy 

... smx-ainy = tan.fr-1 = tan (> - 45°) 

sin a; + sin ?/ tan </> + 1 

[(14) of Art. 61]. 

,\ tan ^ (x — y) = tan ^ (x + y) tan (<£ — 45°) 

[(13) of Art. 61]. 

= tan (45° - <£) tan \ (a + j8 + C) . (2) 

thns from (1) and (2) # and ?/ are fonnd. 

EXAMPLES. 

Solve the following right triangles : 






1. Given 


a = 51.303, 


c = 150; 




find 


A =20°, 


B = 70°, 


6=140.95. 


2. Given 


a= 157.33, 


c=250; 




find 


A =39°, 


B=51°, 


6=194.28. 



3. Given a =104, c=185; 

find A=34°12'19".6, B = 55° 47' 40 ".4, 6=153. 

4. Given a =304, c=425; 

find A=45°40' 2".3, B=44°19'57".7, 6=297. 

5. Given 6=3, c=5; 

find A=53° 7'48".4, B=36°52'll".6, a=4. 

6. Given 6 = 15, c=17; 

find A =28° 4'20".9, B = 61°55'39".l, a =8. 

7. Given 6 = 21, c=29; 

find A=43°36'10".l, B=46°23'49".9, a=20. 

8. Given 6 = 7, c=25; 

find A=73°44'23".3, B = 16° 15' 36". 7, a=24. 



19C 


1 


PLANE TRIGONOMETRY. 




9. 


Given 


6=33, 


c=65; 






find 


A=59°29'23".2, 


B=30°30'36".8, 


a =56. 


10. 


Given 


c=625, 


A=44°; 






find 


a =434.161, 


6=449.587. 




11. 


Given 


c=300, 


A=52°; 






find 


a =236.403, 


6=184.698. 




12. 


Given 


c=13, 


A= 67°22'48".5 


'■> 




find 


B=22°37'll".5, 


a=12 


6=5. 


13. 


Given 


A=77°19'10".6, 


c=41; 






find 


B=12°40'49".4, 


a=40, 


6=9. 


14. 


Given 


B=48°53'16".5, 


c=73; 






find 


A=41° 6'43".5, 


a=48, 


6=55. 


15. 


Given 


B = 64° 0'38".8, 


c=89; 






find 


A=25°59'21".2, 


a=39, 


6=80. 


16. 


Given 


A=77°19'10".6, 


a=40; 






find 


B=12°40'49".4, 


6=9, 


c=41. 


17. 


Given 


A=87°12'20".3, 


a=840; 


c=841. 




find 


B= 2°47'39".7, 


6=41, 


18. 


Given 


A=32°31'13".5, 


a=336; 






find 


B=57°28'46".5, 


6=527, 


c=625. 


19. 


Given 


A =82° 41' 44", 


a=1100; 






find 


B= 7° 18' 16", 


6=141, 


c=1109. 


20. 


Given 


A=75°23'18".5, 


6=195; 






find 


B=14°36'41".5, 


a =748, 


c=773. 


21. 


Given 


B = 87° 49' 10", 


6=42536.37; 






find 


A= 2° 10' 50", 


a=1619.626, 


c= 42567.2. 






EXAMPLES. 191 

22. Given A= 88° 59' 6=2.234875; 

find B = 1° V, o= 125.9365, c= 125.9563. 

23. Given A= 35°16'25", a=388.2647; 

find B= 54°43'35," 6=548.9018, c=672.3412. 

24. Given a = 7694.5, 6 = 8471; 

find A= 42°15', B=47°45', c=11444. 

25. Given a = 736, 6=273; 

find A= 69°38'56".3, B=20°21'3".7, c=785. 

26. Given a =200, 6=609; 
find A= 18°10'50", B=71°49'10", c=641. 

27. Given a =276, 6=493; 
find A= 29°14'30".3, B= 60° 45' 29. "7, c=565. 

28. Given a= 396, 6=403; 
find A= 44°29'53", B= 45°30' 7' 1 , c=565. 

29. Given a=278.3, 6=314.6; 

find A= 41° 30', B= 48°30', c=420. 

30. Given a=372, 6 = 423.924; 

find A= 41°16' 2".7, B = 48°43'57".3, c=564. 

31. Given a =526.2, 6=414.745; 

find A= 51°45'18".7, B= 38°14'41".3, c=670. 

Solve tlie following oblique triangles : 



32. 


Given B = 


50° 30', 


C = 122° 9', 


a = 90; 




find A= 


7° 21', 


6=542.850, 


c=595.638. 


33. 


Given A = 


82° 20', 


B= 43° 20', 


a=479 ; 




find C = 


54° 20', 


6=331.657, 


c=392.473. 


34. 


Given A= 


79° 59', 


B= 44° 41', 


a=795 ; 




find C = 


51° 20', 


6=567.888, 


c= 663.986. 



192 PLANE TRIGONOMETBY. 



35. 


Given B = 


37° 58', • 


C = 65° 2', 


a=999; 




find A= 


77° 0', 


6=630.771, 


c= 829.480. 


36. 


Given A= 


70° 55', 


C= 52° 9', 


a=6412 ; 




find B= 


56° 56', 


6=5686.00, 


c= 5357.50. 


37. 


Given A= 


48° 20', 


B= 81° 2' 16", 


5=5.75; 




find C= 


50° 37' 44", 


a=4.3485, 


c=4.5. 


38. 


Given A= 


72° 4', 


B = 41° 56' 18", 


c=24; 




find C = 


65° 59' 42", 


a =24.995, 


6 = 17.559. 



39. Given A= 43°36'10".l, C=124°58'33".6, 6=29; 

find B= 11°25'16".3, o=101, c=120. 

40. Given A= 69° 59' 2".5, C= 70°42'30", 6=149; 

find B= 39°18'27".5, a=221, c=222. 

41. Given A= 21° 14' 25", a=345, . 6=695; 

find B= 46° 52' 10", C= 111° 53' 25", c= 883.65. 

or B'=133° 7' 50", C'= 25°37'45", c'=411.92. 

42. Given A= 41° 13' 0", a=77.04, 6=91.06; 

find B= 51° 9' 6", C= 87° 37' 54", c= 116.82, 

or B'=128°50'54", C'= 9°56' 6", c'=20.172. 

43. Given A = 21° 14' 25", a =309, 6=360; 

find B= 24°57'54", C=133°47'41", c=615.67, 

or B'=155° 2' 6", C'= 3°43'29", c'=55.41. 

44. Given B= 68°10'24", a=83.856, 6=85.153; 

find A= 66° 5' 19", C= 45° 44' 17", c= 65.696. 

45. Given B= 60° 0'32", a=27.548, 6=35.055; 

find A= 42 D 53'34", C= 77° 5'54", c=39.453. 

46. Given A = 60°, a =120, 6 = 80; 

find B= 35°15'52", C= 84°44' 8", c=137.9796. 



EXAMPLES. 193 

47. Given A = 50°, a =119, 6=97; 

find B= 38°38'24", C= 91°21'36", c=155.3. 

48. Given C= 65° 59', a=25, c=24; 

find A= 72° 4' 48", B= 41°56'12", 6=17.56, 
or A'=107°55'12", B'= 6° 5' 48". 

49. Given A = 18°55'28".7, a=13, 6 = 37; 

find B= 67°22'48".l, or B'= 112°37'11".9. 

50. Given C= 15°11'21", a=232, 6=229; 

find A= 85°11'58", B= 79°36'40", c=61. 

51. Given C = 126°12'14", a=5132, 6=3476; 

find A= 32° 28' 19", B= 21°19'27", c=7713.3. 

52. Given C= 55°12' 3", a=20.71, 6=18.87; 

find A= 67° 28' 51 ".5, B= 57°19'5".5, c= 18.41. 

53. Given C= 12°35' 8", a =8.54, 6=6.39; 

find A=136°15'48", B= 31° 9' 4", c=2.69. 

54. Given C= 34° 9'16", a=3184, 6=917; 

find A=133°51'34", B= 11°59'10", c=2479.2. 

55. Given C= 32°10'53".8, a=101, 6=29; 

find A=136°23'49".9, B= 11°25'16".3, c=78. 

56. Given C= 96°57'20".l, a=401, 6=41; 

find A= 77°19'10".6, B= 5°43'29".2, c=408. 

57. Given C= 30° 40' 35", a =221, 6 = 149; 

find A=110° 0'57".5, B= 39°18'27".5, c=120. 

58. Given C= 66°59'25".4, a=109, 6=61; 

find A= 79° 36' 40", B= 33° 23' 54". 6, c=102. 

59. Given C = 131° 24' 44", a =229, 6=109; 

find A= 33°23'54".6, B= 15°11'21".4, c=312. 



194 PLANE TRIGONOMETRY. 

60. Given C= 104° 3' 51", a =241, 6=169; 

find A=45°46'16".5,B=30° 9'52".5, c=332.97. 

61. Given a= 289, 6=601, c=712; 

find A =23° 32' 12", B=56° 8' 42", C=100°19' 6". 

62. Given a=17, 6=113, c=120; 

findA= 7°37'42", B=61°55'38", C = 110°26'40". 

63. Given a =15.47, 6=17.39, c=22.88; 

find A=42°30'44", B=49°25'49", C=88°3'27". 

64. Given a= 5134, 6=7268, c=9313; 

find A =33° 15' 39", B=50°56' 0", C=95°48'21". 

65. Given a= 99, 6=101, c=158; 

findA=37°22'19", B=38°15'41", C=104°22'0". 

66. Given a=ll, 6=13, c=16; 

findA=43° 2'56", B=53°46'44", C=83°10'20". 

67. Given a=25, 6=26, c=27; 

findA=56°15' 4", B=59°51'10", C = 63°53'46". 

68. Given a= 197, 6=53, c=240; 

find A=31°53'26".8,B= 8°10'16".4, C = 139°56'16".8. 

69. Given a = 509, 6 = 221, c = 480 ; 

find A=84°32'50".5,B=25°36'30".7, C=69°50'38".8. 

70. Given a= 533, 6=317, c=510; 

findA=76°18'52", B=35°18' 0".9, C=68°23'7".l. 

71. Given a= 565, 6=445, c=606; 

find A=62°51'32".9,B=44°29'53", C=72°38'34".l. 

72. Given a=10, 6=12, c=14; 

find A=44°24'55".2,B=57° 7'18", C = 78°27'47". 

73. Given a=.8706, 6=.0916, c=.7902; 

find A=149°49'0".4,B= 3° 1'56".2, C=27°9'3".4. 






, 



EXAMPLES. 



195 



Find the area : 



74. G 

75. 
76. 
77. 
78. 
79. 
80. 
81. 
82. 
83. 
84. 
85. 
86. 



ven a =10, 
a =40, 
6 = 7, 



6=12, C = 60°. 
6=60, C=30°. 
c=5V2, A=135° 



Ans. 30V3. 
600. 

a=32.5, 6=56.3, C=47°5'30". 670. 

6=149, A= 70° 42' 30", B= 39° 18' 28"." 15540. 

c=8.025, B=100° 5'23", C=31° 6'12". 46.177. 

a=5, 6=6, c=7. 12. 

a=625, 6=505, c=904. 151872. 

a=409, 6=169, c=510. 30600. 

a=577, 6=73, c=520. 12480. 

a=52.53, 6=48.76, c=44.98. 1016.9487. 

a=13, 6=14, c=15. 84. 

a =242 yards, 6=1212 yards, c=1450 yards. 

Ans. 6 acres. 



87. " a=7.152, 6=8.263, c=9.375. 



28.47717. 



88. The sides of a triangle are as 2 : 3 : 4 : show that the 
radii of the escribed circles are as \ : ^ : 1. 

89. The area of a triangle is an acre ; two of its sides 
are 127 yards and 150 yards : find the angle between them. 

Ans. 30° 32' 23". 

90. The adjacent sides of a parallelogram are 5 and 8, 
and they include an angle of 60° : find (1) the two diag- 
onals, and (2) the area. Ans. (1) 7, Vll!9 ; (2) 20 V3. 

91. Two angles of a triangular field are 22^-° and 45°, and 
the length of the side opposite the latter is a furlong. Show 
that the field contains 1\ acres. 



196 PLANE TRIGONOMETRY. 



HEIGHTS AND DISTANCES. 

92. At a point 200 feet in a horizontal line from the foot 
of a tower, the angle of elevation of the top of the tower is 
observed to be 60° : find the height of the tower. 

Arts. 346 feet. 

93. From the top of a vertical cliff, the angle of depres- 
sion of a point on the shore 150 feet from the base of the 
cliff, is observed to be 30° : find the height of the cliff. 

Ans. 86.6 feet. 

94. From the top of a tower 117 feet high, the angle of 
depression of the top of a house 37 feet high is observed to 
be 30° : how far is the top of the house from the tower ? 

Ans. 138.5 feet. 

95. The shadow of a tower in the sunlight is observed 
to be 100 feet long, and at the same time the shadow of a 
lamp-post 9 feet high is observed to be 3V3 feet long : find 
the angle of elevation of the sun, and height of the tower. 

Ans. 60°; 173.2 feet. 

96. A flagstaff 25 feet high stands on the top of a 
house; from a point on the plain on which the house 
stands, the angles of elevation of the top and bottom of 
the flagstaff are observed to be 60° and 45° respectively : 
find the height of the house above the point of observation. 

Ans. 34.15 feet. 

97. From the top of a cliff 100 feet high, the angles of 
depression of two ships at sea are observed to be 45° and 
30° respectively ; if the line joining the ships points directly 
to the foot of the cliff, find the distance between the ships. 

Ans. 73.2. 

98. A tower 100 feet high stands on the top of a cliff ; 
from a point on the sand at the foot of the cliff the angles 



EXAMPLES. 19T 

of elevation of the top and bottom of the tower are observed 
to be 75° and 60° respectively : find the height of the cliff. 

Ans. 86.6 feet. 

99. A man walking a straight road observes at one mile- 
stone a house in a direction making an angle of 30° with 
the road, and at the next milestone the angle is 60° : how 
far is the house from the road ? Ans. 1524 yds. 

100. A man stands at a point A on the bank AB of a 
straight river and observes that the line joining A to a post 
C on the opposite bank makes with AB an angle of 30°. 
He then goes 400 yards along the bank to B and finds that 
BC makes with BA an angle of 60° : find the breadth of 
the river. Ans. 173.2 yards. 

101. From the top of a hill the angles of depression of 
the top and bottom of a flagstaff 25 feet high at the foot 
of the hill are observed to be 45° 13' and 47° 12' respectively : 
find the height of the hill. Ans. 373 feet. 

102. From each of two stations, east and west of each 
other, the altitude of a balloon is observed to be 45°, and 
its bearings to be respectively IST.W. and IsT.E. ; if the sta- 
tions be 1 mile apart, find the height of the balloon. 

Ans. 3733 feet. 

103. The angle of elevation of a balloon from a station 
due south of it is 60°, and from another station due west 
of the former and distant a mile from it is 45° : find the 
height of the balloon. Ans. 6468 feet. 

104. Find the height of a hill, the angle of elevation at 
its foot being 60°, and at a point 500 yards from the foot 
along a horizontal plane 30°. Ans. 250V3 yards. 

105. A tower 51 feet high has a mark at a height of 25 
feet from the ground : find at what distance from the foot 
the two parts subtend equal angles. 



198 PLANE TRIGONOMETRY. 

106. The angles of a triangle are as 1 : 2 : 3, and the per- 
pendicular from the greatest angle on the opposite side is 
30 yards : find the sides. Ans. 20 V3, 60, 40 V3. 

107. At two points A, B, an object DE, situated in the 
same vertical line CE, subtends the same angle a ; if AC, 
BC be in the same right line, and equal to a and 6, respec- 
tively, prove 

DE = (a + 6) tan a. 

108. From a station B at the foot of an inclined plane 
BC the angle of elevation of the summit A of a mountain is 
60°, the inclination of BC is 30°, the angle BCA 135°, and 
the length of BC is 1000 yards : find the height of A over B. 

Ans. 500(3 + V3) yards. 

109. A right triangle rests on its hypotenuse, the length 
of which is 100 feet ; one of the angles is 36°, and the 
inclination of the plane of the triangle to the horizon is 
60°: find the height of the vertex above the ground. 

Ans. 25 V3 cos 18°. 

110. A station at A is due west of a railway train at B ; 
after traveling N.W. 6 miles, the bearing of A from the 
train is S. 22^° W. : find the distance AB. Ans. 6 miles. 

111. The angles of depression of the top and bottom of a 
column observed from a tower 108 feet high are 30° and 60° 
respectively : find the height of the column. Ans. 72 feet. 

112. At the foot of a mountain the elevation of its sum- 
mit is found to be 45°. After ascending for one mile, at a 
slope of 15°, towards the summit, its elevation is found to 
be 60° : find the height of the mountain. 

Ans. ^— miles. 

V2 

113. A and B are two stations on a hillside. The inclina- 
tion of the hill to the horizon is 30°. The distance between 
A and B is 500 yards. C is the summit of another hill in 



.. 



EXAMPLES. 199 

the same vertical plane as A and B, on a level with A, but 
at B its elevation above the horizon is 15°: find the distance 
between A and C. Ans. 500 ( V3 + 1) . 

114. From the top of a cliff the angles of depression of 
the top and bottom of a lighthouse 97.25 feet high are 
observed to be 23° 17' and 24° 19' respectively : how much 
higher is the cliff than the lighthouse ? Ans. 1942 feet. 

115. The angle of elevation of a balloon from a station 
due south of it is 47° 18' 30", and from another station due 
west of the former, and distant 671.38 feet from it, the 
elevation is 41° 14' : find the height of the balloon. 

Ans. 1000 feet. 



116. A person standing on the bank of a river observes 
he elevation of the top of a tree on the opposite bank to be 
51° ; and when he retires 30 feet from the river's bank he 
observes the elevation to be 46° : find the breadth of the 
river. Ans. 155.823 feet. 



117. From the top of a hill I observe two milestones on 
the level ground in a straight line before me, and I find 
their angles of depression to be respectively 5° and 15° : 
find the height of the hill. Ans. 228.6307 yards. 

118. A tower is situated on the top of a hill whose angle 
of inclination to the horizon is 30°. The angle subtended 
by the tower at the foot of the hill is found by an observer 
to be 15° ; and on ascending 485 feet up the hill the tower 
is found to subtend an angle of 30° : find (1) the height of 
the tower, and (2) the distance of its base from the foot of 
the hill. Ans. (1)280.015; (2) 765.015 feet. 

119. The angle of elevation of a tower at a place A due 
south of it is 30°; and at a place B, due west of A, and at 
a distance a from it, the elevation is 18° : show that the 

height of the tower is — - 

V2 + 2V5 



200 PLANE TRIGONOMETRY. 

120. On the bank of a river there is a column 200 feet 
high supporting a statue 30 feet high. The statue to an 
observer on the opposite bank subtends an equal angle with 
a man 6 feet high standing at the base of the column : find 
the breadth of the river. Ans. 10V115 feet. 

121. A man walking along a straight road at the rate of 
3 miles an hour, sees in front of him, at an elevation of 60°, 
a balloon which is travelling horizontally in the same direc- 
tion at the rate of 6 miles an hour; ten minutes after he 
observes that the elevation is 30° : prove that the height of 
the balloon above the road is 440 V3 yards. 

122. An observer in a balloon observes the angle of 
depression of an object on the ground, due south, to be 
35° 30'. The balloon drifts due east, at the same elevation, 
for 2\ miles, when the angle of depression of the same 
object is observed to be 23° 14': find the height of the 
balloon. Ans. 1.34394 miles. 

123. A column, on a pedestal 20 feet high, subtends an 
angle 45° to a person on the ground; on approaching 20 
feet, it again subtends an angle 45°: show that the height 
of the column is 100 feet. 

124. A tower 51 feet high has a mark 25 feet from the 
ground : find at what distance the two parts subtend equal 
angles to an eye 5 feet from the ground. Arts. 160 feet. 

125. From the extremities of a sea-wall, 300 feet long, 
the bearings of a boat at sea were observed to be N. 23° 30' 
E., and TSL 35° 15' W. : find the distance of the boat from 
the sea-wall. Ans. 262.82 feet. 

126. ABC is a triangle on a horizontal plane, on which 
stands a tower CD, whose elevation at A is 50° 3' 2" ; AB is 
100.62 feet, and BC and AC make with AB angles 40° 35' 17" 
and 9° 59' 50" respectively : find CD. Ans. 101.166 feet. 



EXAMPLES. 201 

127. The angle of elevation of a tower at a distance of 
20 yards from its foot is three times as great as the angle 
of elevation 100 yards from the same point: show that the 

height of the tower is — - feet. 

V7 

128. A man standing at a point A, due south of a tower 
built on a horizontal plain, observes the altitude of the tower 
to be 60°. He then walks to a point B due west from A 
and observes the altitude to be 45°, and then at the point C 
in AB produced he observes the altitude to be 30° : prove 
that AB = BC. 

129. The angle of elevation of a balloon, which is ascend- 
ing uniformly and vertically, when it is one mile high is 
observed to be 35° 20' ; 20 minutes later the elevation is 
observed to be 55° 40' : how fast is the balloon moving ? 

Ans. 3 (sin 20° 20') (sec 55° 40') (cosec 35° 20') miles per hour. 

130. The angle of elevation of the top of a steeple at a 
place due south of it is 45°, and at another place due west 
of the former station and distant 100 feet from it the 
elevation is 15°: show that the height of the steeple is 
50(3* -3"*) feet. 

131. A tower stands at the foot of an inclined plane 
whose inclination to the horizon is 9°; a line is measured 
up the incline from the foot of the tower, of 100 feet in 
length. At the upper extremity of this line the tower sub- 
tends an angle of 54° : find the height of the tower. 

Ans. 114.4 feet. 

132. The altitude of a certain rock is observed to be 47°, 
and after walking 1000 feet towards the rock, up a slope 
inclined at an angle of 32° to the horizon, the observer finds 
that the altitude is 77°: prove, that the vertical height of 
the rock above the first point of observation is 1034 feet. 



202 PLANE TRIGONOMETRY. 

133. From a window it is observed that the angle of 
elevation of the top of a house on the opposite side of the 
street is 29°, and the angle of depression of the bottom of 
the house is 56°: find the height of the house, supposing 
the breadth of the street to be 80 feet. Ans. 162.95 feet. 

134. A and B are two positions on opposite sides of a 
mountain ; C is a point visible from A and B ; AC and BC 
are 10 miles and 8 miles respectively, and the angle BCA is 
60°: prove that the distance between A and B is 9.165 
miles. 

135. P and Q are two inaccessible objects ; a straight line 
AB, in the same plane as P and Q, is measured, and found 
to be 280 yards ; the angle PAB is 95°, the angle QAB is 
47^°, the angle QBA is 110°, and the angle PBA is 52° 20' : 
find the length of PQ. Ans. 509.77 yards. 

136. Two hills each 264 feet high are just visible from 
each other over the sea : how far are they apart ? (Take 
the radius of the earth = 4000 miles.) Ans. 40 miles. 

137. A ship sailing out of harbor is watched by an 
observer from the shore ; and at the instant she disappears 
below the horizon he ascends to a height of 20 feet, and 
thus keeps her in sight 40 minutes longer : find the rate at 
which the ship is sailing, assuming the earth's radius to be 
4000 miles, and neglecting the height of the observer. 

Ans. 40V330 feet per minute. 

138. From the top of the mast of a ship 64 feet above 
the level of the sea the light of a distant lighthouse is just 
seen in the horizon ; and after the ship has sailed directly 
towards the light for 30 minutes it is seen from the deck 
of the ship, which is 16 feet above the sea : find the rate 
at which the ship is sailing. (Take radius = 4000 miles.) 

Ans. 8Vff miles per hour. 



EXAMPLES. 203 

139. A, B, C, are three objects at known distances apart ; 
namely, AB = 1056 yards, AC = 924 yards, BC = 1716 
yards. An observer places himself at a station P from 
which C appears directly in front of A, and observes the 
angle CPB to be 14° 24' : find the distance CP. 

Ans. 2109.824 yards. 

140. A, B, C, are three objects such that AB = 320 yards, 
AC = 600 yards, and BC = 435 yards. From a station P 
it is observed that APB = 15°, and BPC = 30° : find the 
distances of P from A, B, and C ; the point B being near- 
est to P, and the angle APC being the sum of the angles 
APB and BPC. Ans. PA = 777, PB = 502, PC = 790. 



204 PLANE TRIGONOMETRY. 



CHAPTER VIII. 

CONSTKUCTION OF LOGAEITHMIG AND TKIG0N0METKIC 

TABLES. 

128. Logarithmic and Trigonometric Tables. — In Chap- 
ters IV., V., and VII., it was shown how to use logarithmic 
and trigonometric tables ; it will now be shown how to 
calculate such tables. Although the trigonometric func- 
tions are seldom capable of being expressed exactly, yet 
they can be found approximately for any angle; and the 
calculations may be carried to any assigned degree of accu- 
racy. We shall first show how to calculate logarithmic 
tables, and shall repeat here substantially Arts. 208, 209, 
210, from the College Algebra. 

129. Exponential Series. — To expand e x in a series of 
ascending poivers of x. 

By the Binomial Theorem, 



U , A 1 , 1 , nx(nx-l) 1 
\ nj n [2 n 2 



nx(nx — 1) (nx — 2) 1 



n° 



xfx ] x( X )( X—-) 

1+X+ X ?7 ii + V »A »; + ... (1) 



12 

Similarly, 



n 



i_i fi-lVi- 2 



i + i 



L 



\2 ' [3 



(2) 



LOGARITHMIC SERIES. 205 

and therefore series (1) is equal to series (2) however great 
n may be. Hence if n be indefinitely increased, we have 
from (1) and (2) 

i+ * + i + I + ii + -=( i+i+ i + H + - 

The series in the parenthesis is usually denoted by e; 
hence e* = l + x +£ + £ + £+ (3) 

If 12 li 

which is the expansion of e x in powers of x. 
This result is called the Exponential Theorem. 
If we put x = 1, we have from (3) 

From this series we may readily compute the approxi- 
mate value of e to any required degree of accuracy. This 
constant value e is called the Napierian base (Art. 64). To 
ten places of decimals it is found -to be 2.7182818284. 

Cor. Let a = e c ; then c = log e a, and a x = e cx . Substi- 
tuting in (3), we have 

C4/y*6 />3/ywJ 

or a«=l + *log;d+^^ 2 + ^5|^+... (4) 

which is the expansion of a x in powers of x. 

130. Logarithmic Series. — To expand log e (l + a?) in a 
series of ascending powers of x. 
By the Binomial Theorem, 
a x = (1 + a - l) x = 1 + x (a - 1) + x < x ~ 1 ) ( a _ iy 

+ »(s-l)(s-2) (a _ 1)8+ _ 



206 PLANE TBIGONOMETBY. 

- 1 + x [a _i _ i( a - iy+ \{a - l) 3 - i(a-l) 4 + •••] 
+ terms involving x 2 , X s , etc. 

Comparing this value of a x with that given in (4) of Art. 
129, and equating the coefficients of x, we have 

log e a = a-l-|(a-l) 2 + i(a-l) 3 -i(a-l) 4 +... 

Put a = 1 + x ; then 

log;(l + A) = a ; -|.+ |-| + (3) 

This is the Logarithmic Seines; but unless # be very 
small, the terms diminish so slowly that a large number of 
them will have to be taken ; and hence the series is of little 
practical use for numerical calculation. If x > 1, the series 
is altogether unsuitable. We shall therefore deduce some 
more convenient formulae. 

Changing x into — x, (1) becomes 

X" X X 

log.(l-fls) = -a------- (2) 

Subtracting (2) from (1), we have 

log .l±! = 2(Wf + f + f + ...) ... (3) 
-r, , 1 +x n + 1 

Put — - 1 = ! .'• %=■ 



1 — x n 2n + 1' 

and (3) becomes 
1 n-\-l o 

71 



-]• 



or log e O + 1) 
=log e ?i + 2 



2n + l 3(2?i + l) 3 5(2n + l) 5 

f ../J . (4) 



v r > 



2n + l 3(2n + l) 3 5(2n + l) 5 

This series is rapidly convergent, and gives the logarithm 
of either of two consecutive numbers to any extent when 
the logarithm of the other number is known. 



COMPUTATION OF LOGARITHMS. 



207 



131. Computation of Logarithms. — Logarithms to the 
base e are called Napierian Logarithms (Art. 64). They 
are also called natural logarithms, because they are the 
first logarithms which occur in the investigation of a 
method of calculating logarithms. Logarithms to the base 
10 are called common logarithms. When logarithms are 
used in theoretical investigations, the base e is always 
understood, just as in all practical calculations the base 
10 is invariably employed. It is only necessary to com- 
pute the logarithms of prime numbers from the series, 
since the logarithm of a composite number may be obtained 
by adding together the logarithms of its component factors. 
The logarithm of 1 = 0. Putting n = 1, 2, 4, 6, etc., suc- 
cessively, in (4) of Art. 130, we obtain the following 



Napierian Logarithms : 



log e 2 = 2 



+ 



1 +A.+ * 



log e 3=log e 2+2 
log, 4 = 21og e 2 
log, 5=log e 4+2 



3 3 • 3 3 5 • 3 5 7 • 3 7 9 • 3 9 
1,1,1,1 



+ 



5 3 • 5 3 5 • 5 5 7 • 5 7 



*-h * 



_9 3 • 9 3 5 • 9 5 7 • 9 7 
log e 6 = log e 2 + log e 3 

1 



log e 7 = log e 6 + 2 

log e 8 = 31og e 2 
log e 9 = 21og e 3 
log e 10 = log e 5 + log e 2 
And so on. 



* + 1 



13 3-13 3 5-13 5 



= 0.69314718. 

= 1.09861228. 

= 1.38629436. 

= 1.60943790. 

= 1.79175946. 

= 1.94590996. 

= 2.07944154. 
= 2.19722456. 
= 2.30258509. 



The number of terms of the series which it is necessary 
to include diminishes as n increases. Thus, in computing 



208 



PLANE TRIGONOMETRY. 



the logarithm of 101, the first term of the series gives 
the result true to seven decimal places. 

By changing b to 10 and a to e in (1) of Art. 65, we have 

log e m __ log e m 



log 10 m: 



.43429448 log e ra, 



or 



log e 10 2.30258509 
common logm = Napierian logm x .43429448. 

The number .43429448 is called the modulus of the common 
system. It is usually denoted by /a. 

Hence, the common logarithm of any number is equal to 
the Napierian logarithm of the same number multiplied by 
the modulus of the common system, .43429448. 

Multiplying (4) of Art. 130 by /*, we obtain a series by 
which common logarithms may be computed ; thus, 

logioO + l) = 

1,1 1 



log 10 n + 2fx 



_2n + l ' 3(2n + l) 3 + 5(2n + l) 



: + 



-] 



(1) 



Common Logarithms. 
log 10 2 = /xlog e 2 = .43429448 x .69314718 = .3010300. 
log 10 3 = /xlog e 3 = .43429448 x 1.09861228 = .4771213. 
log 10 4 = 21og 10 2 = .6020600. 

log 10 5 = /xlog e 5 = .43429448 x 1.60943790 = .6989700. 
And so on. 

132. If 6 be the Circular Measure of an Acute Angle, 
sin 9, 6, and tan are in Ascending Order of Magnitude. 

With centre O, and any radius, de- 
scribe an arc BAB'. Bisect the angle 
BOB' by OA; join BB', and draw the 
tangents BT, B'T. 

Let AOB = AOB' = ft Then 

BB' < arc BAB' < BT + B'T 

(Geom., Art. 246) 
.-. BC<arcBA<BT. 




LIMITING VALUES OF SIN 6. 209 

BC BA BT 

" OB OB OB* 

.-. sin 6 < 6 < tan 6. 

133. The Limit of ^^-, when 6 is Indefinitely Diminished, 
is Unity. 

We have sin 6 < 6 < tan 6 (Art. 132) 

.-. l<-i-< sec<9. 

Now as is diminished indefinitely, sec 6 approaches the 
limit unity ; then when 6 = 0, we have sec = 1. 

Q 

.-. the limit of , which lies between sec0 and unity, 

sin0 
is unity. 

.\ also approaches the limit unity. 

* tan 6 sin n j.vt m. * tan T A . . 

As = X sec 0, the limit oi , when is m- 

6 6 , 6 

definitely diminished, is also unity. 
This is often stated briefly thus : 

^i = l,aad*^ = l, when = 0. 

Note. — From this it follows that the sines and the tangents of very small 
angles are proportional to the angles themselves. 

134. If 6 is the Circular Measure of an Acute Angle, sin 6 
lies between 6 and 6 — — ; and cos 6 lies between 1 — T and 

2 16' 
(1) We have tan^>^ (Art. 132) 

. 0^ 6 6 

.-. sin- > -cos-- 
2 2 2 



210 PLANE TRIGONOMETRY. 

B 8 

.-. 2 sin -cos- >0cos 2 -- 

2 2 2 

.-. sin 0>o(l — sin 2 -^ 

> /l_l 2N \ . . (Art. 132) 

.-. sin0<0 and >0 — -- 

4 

(2) cos<9 = l-2sin 2 |. 

> i - 2 (-D- 

02 

.-. cos > 1 

2 

Also, sin f>|-l(I)V(l). 

••• C0S ^< 1 - 2 [|-II ■ 

2 16 512 

.-. cos0>l-- and <l_i 2 +— • 
2 2 16 

as 

Note. —It may be proved that sin 0> — — , as follows : 

We have 3 sin - -sin0 = 4sin 3 - (Art. 50) (1) 

3 3 

.'. 3 sin- -sin ^=4sin3 t (by putting? for 0) (2) 

3 2 3 3 2 3 

• 

3 8 i n .£_ 8 i n _JL =4sin3-?- (w) 

371 3 r>-l 3»i 

Multiply (1), (2), ». (n) by 1, 3, — 3 71 "" 1 , respectively, and add them, 

3 n sin 1 - sin = 4(sin3 * + 3 sins - + ... Z"- 1 sin* AV 
3" V 3 32 3V 



SINE AND COSINE OF 10" AND OF l\ 211 



Bill 

.. ,.__._ 8m , <4 (- + - + ,.. ) (Art. 



132) 



3 n 



<l,3/ 1+ i + ..._i_y 

33 \ 3 2 32n-2/ 





If n = 00 , then 


sin — 

-1!-! 






_0 

3 n 






and i*>(l+I + 


1 \_403 1 _ 03 
" Z 2n-2J ~ 33 * j _ 1 ~ 6 
3 2 






... 


0-sin0<-, and .*. sin0>0- 
6 


-ff. 

6* 



(Art. 133) 



This makes the limits for sin closer than in (1) of this Art. 

135. To calculate the Sine and Cosine of 10" and of 1' 

(1) Let be the circular measure of 10". 

Then 

10 it 3.141592653589793... 



= 



-j 



180 x 60 x 60 64800 

or 6 = .000048481368110 ■ ■ -, correct to 15 decimal places. 

.-. ^=. 000000000000032-.., " « " " 

... 0-- = . 000048481368078-.., " " " 

4 

#3 
Hence the two quantities 6 and agree to 12 deci- 

mal places ; and since sin 6 < 6 and > 6 — - (Art. 134), 

.-. sin 10" = .000048481368, to 12 decimal places. 
We have 



cos 10" = VI -sin 2 10" = 1 - i sin 2 10" 

= .9999999988248 •-, to 13 decimal places. 

Or we may use the results established in (2) of Art. 134, 
and obtain the same value. 



212 PLANE TRIGONOMETRY. 

(2) Let be the circular measure of V. 

Then 

6 = — * — - = .000290888208665, to 15 decimal places. 
180 x 60 

... 6 1 =.000000000006 to 12 " " 

4 

... e~ = .00029088820 to 11 « " 

4 

Hence 6 and 6 differ only in the twelfth decimal. 

.-. sin 1' =. .00029088820 to 11 decimal places. 



cos 1' = Vl-s'in 2 l'=.999999957692025 to 15 decimal places, 
Othenvise thus : 

1 -^ = .999999957692025029 to 18 decimal places. 

and — = .00000000000000044 to 17 decimal places. 
16 r 

But cos V > 1 - - and <l-^ + -^ . . . (Art. 134) 

.-. cos V = .999999957692025 to 15 decimal places, as before. 

Cor. 1. The sine o/10" equals the circular measure of 10", 
to 12 decimal places ; and the sine of V equals the circular 
measure ofV to 11 decimal places. 

Cor. 2. If n denote any number of seconds less than 60, we 
shall have approximately 

sin n" = n sin l' f , 

for the sine of n fl = the circular measure of n", approxi- 
mately, = n times the circular measure of 1". 

Cgr g n = circular measure of ^ approximately . that 
is, the number of seconds in any small angle is found 






TABLES OF NATURAL SINES AND COSINES. 213 

approximately by dividing the circular measure of that 
angle by the sine of one second. 

136. To construct a Table of Natural Sines and Cosines at 
Intervals of 1'. 

We have, by Art. 45, 

sin(# + y) = 2 since cosy — sin(x — y), 
cos (x + y) = 2 cos x cos y — cos (x — ?/) . 

Suppose the angles to increase by 1'; putting y = V, we 
have, 

sin (x + 1') = 2 since cos 1' — sin (x — V) .... (1) 

cos(a; + l') = 2cosa;cosl' — cos(# — 1') . . . . (2) 

Putting x=V, 2', 3', 4', etc., in (1) and (2), we get for 
the sines 

sin 2' = 2 sin V cos 1' - sin ; = .0005817764, 

sin 3' = 2 sin 2' cos 1' - sin 1' = .0008726646, 
sin 4' = 2 sin 3' cos 1' - sin 2' = .0011635526 ; 
and for the cosines 

cos 2 f = 2 cos V cos 1' - cos 0' = .9999998308, 
cos 3' = 2 cos 2' cos 1' - cos V = .9999996193, 
cos 4' = 2 cos 3' cos I 1 - cos 2' = .9999993223. 

We can proceed in this manner * until we find the values 
of the sines and cosines of all angles at intervals of l f from 
0° to 30°. 

137. Another Method. 

Let a denote any angle. Then, in the identity, 

sin (n + l)a = 2 sin na cos a — sin (?i — l)a, 
put 2(1 — cos a) = ft, 

and we get 

sin(n + l)a— sin na=sinnu — sin(n — l)cc — k sinna . (1) 

* This method is due to Thomas Simpson, an English geometrician. 



214 PLANE TRIGONOMETRY. 

This formula enables us to construct a table of sines of 
angles whose common difference is a. 

Thus, suppose a = 10", and let n = 1, 2, 3, 4, etc. 

Then 

sin 20" - sin 10" = sin 10" - k sin 10", 

sin 30" - sin 20" = sin 20" - sin 10" - k sin 20", 

sin 40" _ sin 30" = sin 30" - sin 20" - k sin 30", etc. 

These equations give in succession sin 20", sin 30", etc. 
It will be seen that the most laborious part of this work is 
the multiplication of k by the sines of 10", 20", etc., as they 
are successively found. But from the value of cos 10", we 
have 

k = 2 (1 - cos 10") *= .0000000023504, 

the smallness of which facilitates the process. 

In the same manner a table of cosines can be constructed 
by means of the formula, 

cos (n + 1) a — cos na == cos na — cos (n — 1) a — k cos na, 

which is obtained from the identity, 

cos (n + l)a = 2 cos na cos a — cos (n — l)a, 

by putting 2(1 — cos a) = k, as before. 

138. The Sines and Cosines from 30° to 60°.— It is not 
necessary to calculate in this way the sines and cosines of 
angles beyond 30°, as we can obtain their values for angles 
from 30° to 60° more easily by means of the formulae (Art. 
45): 

sin (30° + a) = cos a — sin (30° - a), 

cos (30° + a) = cos (30° - a) - sin a, 

by giving a all values up to 30°. Thus, 

sin 30° V = cos 1' - sin 29° 59 f , 

cos 30° V = cos 29° 59' - sin V, and so on. 



TABLES OF TANGENTS AND SECANTS. 215 

139. Sines of Angles greater than 45°. — When the sines 
of angles up to 45° have been calculated, those of angles 
between 45° and 90° may be deduced by the formula 

sin (45° + a) — sin (45° — a) = V2 sin a . (Art. 45) 

Also, when the sines of angles up to 60° have been found, 
the remainder up to 90° can be found still more easily from 
the formula 

sin (60° + a) - sin (60° - a) = sin a. 

Having completed a table of sines, the cosines are known, 

smce cos a = sin (90° -a). 

Otherwise thus : When the sines and cosines of the angles 
lp to 45° have been obtained, those of angles between 45° 
id 90° are obtained from the fact that the sine of an 
angle is equal to the cosine of its complement, so that it is 
not necessary to proceed in the calculation beyond 45°. 

Note. — A more modern method of calculating the sines and cosines of angles 
is to use series (3) and (4) of Art. 156. 

140. Tables of Tangents and Secants. — To form a table 
of tangents, we find the tangents of angles up to 45°, from 
the tables of sines and cosines, by means of the formula 

, sin a 

tan a = 

cos a 

Then the tangents of angles from 45° to 90° may be 
obtained by means of the identity * 

tan (45° + a) = tan (45° - a) + 2 tan 2 a. 

When the tangents have been found, the cotangents are 
known, since the cotangent of any angle is equal to the 
tangent of its complement. 

A table of cosecants may be obtained by calculating the 
reciprocals of the sines; or they may be obtained more 

* Called Cagnoli's formula. 



216 PLANE TRIGONOMETRY. 

easily from the tables of the tangents by means of the 

formula 

cosec a =# tan ^ + cot a. 

The secants are then known, since the secant of any 
angle is equal to the cosecant of its complement. 

141. Formulae of Verification. — Formulae used to test the 
accuracy of the calculated sines or cosines of angles are called 
Formulae of Verification. 

It is necessary to have methods of verifying from time 
to time the correctness of the values of the sines and 
cosines of angles calculated by the preceding method, since 
any error made in obtaining the value of one of the func- 
tions would be repeated to the end of the work. For this 
purpose we may compare the value of the sine of any angle 
obtained by the preceding method with its value obtained 
independently. 

V5 — 1 
Thus, for example, we know that sin 18° = — (Art. 

57) ; hence the sine of 18° may be calculated to any degree 
of approximation, and by comparison with the value obtained 
in the tables, we can judge how far we can rely upon the 
tables. 

Similarly, we may compare our results for the angles 
221°, 30°, 36°, 45°, etc., calculated by the preceding method 
with the sines and cosines of the same angles as obtained 
in Arts. 26, 27, 56, 57, 58, etc. 

There are, however, certain well-known formulae of veri- 
fication which can be used to verify any part of the calcu- 
lated tables ; these are 

Euler's Formulae : 
sin (36° + A) - sin (36° - A) + sin (72° - A) 

- sin (72° + A) = sin A. 
cos (36° + A) + cos (36° - A) - cos (72° + A) 

- cos (72° - A) = cos A. 



LOGARITHMIC TRIGONOMETRIC FUNCTIONS. 217 

Legendre's Formula : 
sin (54° + A) + sm (54° - A) - sin (18° + A) 

- sin (18° - A) = cos A. 

The verification consists in giving to A any value, and 
taking from the tables the sines and cosines of the angles 
involved : these values must satisfy the above equations. 

To prove Eulefs Formulae : 
sin (36° + A) - sin (36° - A) = 2 cos 36° sin A . (Art. 45) 

= V5 + l ginA m ( Art58 ) 



sin (72° + A) - sin (72° - A) = 2 cos 72° sin A . (Art. 45) 

= V5-l sinA ^ (Art. 57) 

Subtracting the latter from the former, we get sin A. 
Similarly, Euler's second formula may be proved. 
By substituting 90° — A. for A in this formula we obtain 
Legendre's Formula. 

142. Tables of Logarithmic Trigonometric Functions. — 

To save the trouble of referring twice to tables — first to 
the table of natural functions for the value of the function, 
and then to a table of logarithms for the logarithm of that 
function — it is convenient to calculate the logarithms of 
trigonometric functions, and arrange them in tables, called 
tables of logarithmic sines, cosines, etc. 

When tables of natural sines and cosines have been con- 
structed, tables of logarithmic sines and cosines may be 
made by means of tables of ordinary logarithms, which will 
give the logarithm of the calculated numerical value of the 
sine or cosine of any angle ; adding 10 to the logarithm so 
found we have the corresponding tabular logarithm. The 
logarithmic tangents may be found by the relation 
log tan A — 10 + log sin A — log cos A ; 
and thus a table of logarithmic tangents may be constructed. 



218 PLANE TRIGONOMETRY. 



PROPORTIONAL PARTS. 



143. The Principle of Proportional Parts. — It is often 
necessary to find from a table of logarithms, the logarithm 
of a number containing more digits than are given in the 
table. In order to do this, we assumed, in Chapter IV., the 
principle of proportional parts, which is as follows : 

The differences between three numbers are proportional to the 
corresponding differences between their logarithms, provided 
the differences between the numbers are small compared with 
the numbers. 

By means of this principle, we are enabled to use tables 
of a more moderate size than would otherwise be necessary. 

We shall now investigate how far, and with what excep- 
tions, the principle or rule of proportional increase is true. 

144. To prove the Rule for the Table of Common Loga- 
rithms. 

We have 

log (n + d) — logn = log 71 "*" == log ( 1 + - } 

n \ nj 

-'(r£ + 5?—) • <Artl30) 

where //, = .43429448 •••, a quantity < }. 

Now let n be an integer not < 10000, and d not > 1 ; 

then - is not greater than .0001. 

n 



" 2n 2 



is not >i(.0001) 2 , i.e., not > .0000000025 ; 



and ^— is much less than this. 
3n 3 

.'. log (n + d) — log n = fx~, correct at least as far as seven 
decimal places. n 



RULE OF PROPORTIONAL PARTS. 219 

Hence if the number be changed from n to n + d, the 
corresponding change in the logarithm is approximately 
fid 
n 

Therefore, the change of the logarithm is approximately 
proportional to the change of the number. 

145. To prove the Rule for the Table of Natural Sines. 

sin (0 + h) — sin — sin h cos — sin (1 — cos h) 

= sin ft cos 0( 1 — tan tan- j . (Art. 51) 

If h is the circular measure of a very small angle, sin h = h 
nearly, and tan - = - nearly. 

.\ sin (0 + h) — smO = h cos 0(1 — tan tan - j 

h 2 
= h cos sin 0. 

2 

If ft'is the circular measure of an angle not >V, then 

h is not > .0003 (Art. 135). .-. - is not > .00000005; and 
sin is not > 1. 

.-. sin(0+7i) —sin 0=h cos 0, as far as seven decimal places, 
which proves the proposition. 

Similarly, sin (0 — h) — sin = — h cos 0, approximately. 

146. To prove the Rule for a Table of Natural Cosines. 

cos (0 — h) — cos = sin h sin — cos (1 — cos h) 

= sin h sin 0(1 — cot tan - j- 

If h is the circular measure of a very small angle, sin h = h 
nearly, and tan - = - nearly. 

.*. cos (0 — h) — cos0 = h sin 0(1 — cot tan -J 

h 2 
= h sin0 — — cos0. 



220 PLANE TRIGONOMETRY. 

We may prove, as in Art. 145, that 

7,2 

- cos 6 is not > .00000005. 

2 

.-. cos (6 — ft) — cos = ft sin 6, as far as seven decimal 
places, which proves the proposition. 

Similarly, cos (6 + ft) — cos = — ft sin 0, approximately. 

147. To prove the Rule for a Table of Natural Tangents. 

4an(fl + /0-tanfl = sin ^ + / *> -^= 5HL* 

' cos (0 + ft) cos 6 cos (0 + ft) cos 

tan ft 



cos 2 (1 — tan 6 tan ft) 

If A is the circular measure of a very small angle, 
tan ft = ft nearly. 

.-. tan(0 + ft)-tan0 = hB ^° 
K ' l-fttan0 

== ft sec 2 + ft 2 sin sec 3 0. 

.\ tan (0 + ft) — tan = ft sec 2 0, approximately, 

unless sin sec 3 is large, which proves the proposition. 

Similarly, cot (0 — ft) — cot = ft cosec 2 0, approximately. 

Sc7i. 1. If ft is the circular measure of an angle not > l f , 
then ft is not > .0003. Hence the greatest value of ft 2 sin 

sec 3 is not > .00000009 sin sec 3 0. Therefore, when > -, 

we are liable to an error in the seventh place of decimals. 
Hence the rule is not true for tables of tangents calculated 
for every minute, when the angle is between 45° and 90°. 

Sch. 2. Since the cotangent of an angle is equal to the 
tangent of its complement, it follows immediately that the 
rule must not be used for a table of cotangents, calculated 
for every minute, when the angle lies between 0° and 45°. 



RULE OF PROPORTIONAL PARTS. 221 

148. To prove the Rule for a Table of Logarithmic Sines. 

sin (0 + h) - sin 6 = h cos - - sin . . (Art. 145) 

... 8in (* + ft >=:l + ftcOtfl-g. 

sin 6 2 



.*. log sin (0 + h) — log sin 



_ J _1 f ft cot -A^ 2 + . . .1 (Art. 130) 



2 2 2 



= M ft cot - ^- (1 + cot 2 0) + ••• 
A 

= fxh cot 6 — ^-cosec 2 + ••• 

If ft is the circular measure of an angle not > 10", then 
h is not > .00005, and therefore, unless cot is small or 
cosec 2 large, we Have 

log sin {6 + li) — log sin = fill cot 0, 

as far as seven decimal places, which proves the rule to be 
generally true. 

Sch. 1. When is small, cosec is large. If the log 
sines are calculated to every 10", then h is not >. 00005, 
and /x is not > .5. 

2/1 • * ^ 6 cosec 2 



^fih 2 cosec 2 is not > 



10* 



In order that this error may not affect the seventh decimal 
place, 6 cosec 2 6 must not be > 10 3 , that is, 6 must not be 
less than about 5°. 

When 6 is small, cot 6 is large. Hence, when the angles 



222 PLANE TRIGONOMETRY. 

are small, the differences of consecutive log sines are irregu- 
lar, and they are not insensible. Therefore the rule does 
not apply to the log sine when the angle is less than 5°. 

Sch. 2. When 6 is nearly a right angle, cot is small, 
and cosec approaches unity. 

Hence, when the angles are nearly right angles, the dif- 
ferences of consecutive log sines are irregular and nearly 
insensible. 

149. To prove the Rule for a Table of Logarithmic Co- 
sines. 

h 2 
cos (6 — h) — cos 6 = h sin 6 cos 6 . (Art. 146) 

Li 

., cos(fl-ft) =1 + ft W y 

cos 6 2 



.*. log cos (0 — h) — log cos 6 

= /xlog[l + &tan# j 



: h 



-i^tan(9--Yl 



=s u/itan0 — ^— sec 2 + • •• 
^ 2 

In this case the differences will be irregular and large 
when 6 is nearly a right angle, and irregular and insensible 
when 6 is nearly zero. This is also clear because the sine 
of an angle is the cosine of its complement. 

150. To prove the Rule for a Table of Logarithmic Tan- 
gents. 

tan (<9 + h) - tan 6=h sec 2 + h 2 sin 6 sec 3 6 . (Art. 147) 

.-. tan(fl + /Q =1 + h + A2sec , ft 

tan sin cos 



RULE OF PROPORTIONAL PARTS. 223 

.-. log tan (0 + h) — log tan 



= /* 



- + h 2 sec 2 e--( ^— +7i 2 sec 2 ^Y+...l 

>cos0 2\,sm0cos0 / 



sin cos 

-^ + ,J* 2 (sec 2 0- 



sin 6 cos 6 \ 2 sin 2 cos 2 0/ 

.*. log tan (0 + h) — log tan 

fih _o 7 2 cos 20 
" sin cos /il sin 2 20* 

151. Cases where the Principle of Proportional Parts is 
Inapplicable. 

It appears from the last six Articles that if h is small 
enough, the differences are proportional to h, for values of 
ivhich are neither very small nor nearly equal to a right angle. 

The following exceptional cases arise : 

(1) The difference sin (0 + 7i) — sin is insensible when 
is nearly 90°, for in that case hcosO is very small; it is 
then also irregular, for |-7i 2 sin0 may become comparable 
with h cos 0. 

(2) The difference cos (0 + h) — cos is both insensible 
and irregular when is small. 

(3) The difference tan (0 + h) — tan is irregular when 
is nearly 90°, for ft 2 sin sec 3 may then become compar- 
able with h sec 2 ; it is never insensible, since sec is not 
<1. 

(4) The difference log sin (0 + h) — log sin is irregular 
when is small, and both irregular and insensible when is 
near??/ 90°. 

(5) The difference log cos (0 + h) — log cos is insensible 
and irregular when is small, and irregular when is 
nearly 90°. 

(6) The difference log tan (0 + A) — log tan is irregular 
when is either sraaZZ or nearly 90°. 



224 PLANE TRIGONOMETRY. 

A difference which is insensible is also irregular ; but the 
converse does not hold. 

When the differences for a function are insensible to the 
number of decimal places of the tables, the tables will give 
the functions when the angle is known, but we cannot use 
the tables to find any intermediate angle by means of this 
function; thus, we cannot determine 6 from the value log 
cos 0, for small angles, or from the value log sin 6, for angles 
nearly 90°. 

When the differences for a function are irregular without 
being insensible, the approximate method of proportional 
parts is not sufficient for the determination of the angle by 
means of the function, nor the function by means of the 
angle ; thus, the approximation is inadmissible for log sin 6, 
when is small, for log cos 6, when is nearly 90°, and 
for log tan 6 in either case. (Compare Art. 81.) 

In these cases of irregularity without insensibility, the 
following three means may be used to effect the purpose of 
finding the angle corresponding to a given value of the 
function, or of the function corresponding to a given angle.* 

152. Three Methods to replace the Rule of Proportional 
Parts. 

(1) The simplest plan is to have tables of log sines and 
log tangents, for each second, for the first few degrees of 
the quadrant, and of log cosines and log cotangents, for 
each second, for the few degrees near 90°. Such tables are 
generally given in trigonometric tables of seven places ; 
we can then use the principle of proportional parts for all 
angles which are not extremely near 0° or 90°. 

(2) Delambre's Method. In this method a table is con- 
structed which gives the value of log — — f- log sin 1" for 



every second for the first few degrees of the quadrant. 

* This article has been taken substantially from Hobson's Trigonometry. 



METHODS TO REPLACE THE BULE. 225 

Let be the circular measure of n seconds. Then, when 
is small, we have 6 = n sin 1", approximately. 

t sin# t sin?i" -, „ , , -*,, 

.\ log = log — : — - = log sm n" — log n — log sin 1". 

5 (9 5 nsml" & & 

,\ log sin n" = logn + [log—- — f- log sinl"Y 

Hence, if the angle is known, the table gives the value 
of the expression in parenthesis, and logn can be found 
from the ordinary table of the logs of numbers ; thus log 
sin n" can be found. 

If log sin n" is given, we can find approximately the value 
of n, and then from the table we have the value of the 
expression in parenthesis ; thus we can find log n, and then 
n from an ordinary table of logs of numbers. 

Rem. When 6 is small (less than 5°), 

5HL# = 1 - ^, approximately . . (Art. 134, Note) 

Hence a small error in 6 will not produce a sensible error 

in the result, since log — — will vary much less rapidly 
than 6. 

(3) Maskelyne* s Method. The principle of this method is 
the same as that of Delambre's. If 6 is a small angle, we 
have 

03 

sin 6 = , approximately, 

6 

and cos 6=1-^, « . . (Art. 134) 

. sin^ A W\\ u 



<*-$ 



6 

= (cos (9)*, " 

.-. log sin 6= log 6 + \ log cos 0, approximately. 



226 PLANE TRIGONOMETRY. 

When is a small angle, the differences of log cos are 
insensible (Art. 149) ; hence, if be given, we can find 
log accurately from the table of natural logarithms, and 
also an approximate value of log cos ; the formula then 
gives log sin at once. 

If log sin be given, we must first find an approximate 
value of from the table, and use that for finding log cos 0, 
approximately ; is then obtained from the formula. 

EXAMPLES. 

1. Prove l + 2 + | + i+- -=26. 

2. Prove log- = -- 



2 2 \1 ' 3 • 2 2 ' 5 ' 2 
3. Prove * = 1 + 1 -+1 + 1 + 2 + 3 + 

5. Prove tan0 + £tan 3 + £tan 5 + — 




6. Prove log, 11 = 2.39789527 •• ., by (4) of Art. 130. 

7. Prove log e 13 = 2.56494935 •••, " " 

8. Prove log, 17 = 2.83321334-.., " " 

9. Prove log, 19 = 2.9444394 ..., « « 

10. Pind, by means of the table of common logarithms 
and the modulus, the Napierian logarithms of 1325.07, 
52.9381, and .085623. Ana. 7.18923, 3.96913, - 2.4578. 






EXAMPLES. 227 

Q 

11. Prove that the limit of m sin — is 6, when m = oo. 

m 

12. " a " ratan — is 0, " ra = oo. 

m 

13. « " " — sin— isTrr 2 , " n = oo. 

2 n 

14. " " " Trr 2 tan - is Trr 2 , " n = oo. 

n 

15. " " " vers a ^ is — " 6 = 0. 

vers &# 6 2 ' 

cos -Vis 1, " 71 = 00. 

sin-Yisl, " n = oo. 

cos-Yisl, " n = oo. 

/sin * 
19. " " " [ -|ifll, " n = oo. 

n 



ff\n 2 _02 

20. " " f cos - ] is e *, 



n 



u „ _ ,, 



6\ n • 
21. u " " (cos-) is zero, when n = oo . 



22. If 6 is the circular measure of an acute angle, prove 
(1) cos0<l-| + |^ana (2) tan0>0 + |. 

23. Given — = 1213 th t ^ = 4 o 24 , nearly- 

(9 1014 ^ J 

24. Given ^^ = ?1^: prove that 6 = 3°, nearly. 

2166 F J 



228 PLANE TRIGONOMETRY. 

25. Given sin <£ = n sin 9, tan <j> = 2 tan 6 : find the limit- 
ing values of n that these equations may coexist. 

Ans. n must lie between 1 and 2, or between — 1 and — 2. 

26. Find the limit of (cos ax) coaec2bx , when x = 0. 

27. From a table of natural tangents of seven decimal 
places, show that when an angle is near 60° it may be 
determined within about 2-^- of a second. 

28. When an angle is very near 64° 36 f , show that the 
angle can be determined from its log sine within about -^ 
of a second ; having given (log, 10) tan 64° 36' = 4.8492, and 
the tables reading to seven decimal places. 



BE MOIVRE'S THEOREM. 229 









CHAPTER IX. 

DE MOIVKE'S THEOKEM.* — APPLICATIONS. 

153. De Moivre's Theorem. — For any value of n, positive 
or negative, integral or fractional 

(cos 6 + V^~l sin 6) n = cos n6 + V^l sin nO . . (1) 

I. When n is a positive integer. 
We have the product 
(cos a + V— 1 sin a) (cos ft + V— 1 sin ft) 

= (cos a cos ft— sin a sin ft) + V — l(cos a sin ft+ sin a cos /3) 
= cos (a + /3) +-v /:r T sin (a + /3). 
Similarly, the product 



[cos (a + /3) + V— 1 sin (a + /?)][cosy + V — 1 siny] 
= cos (a + ft + y) + V^T sin (a + ft + y). 

Proceeding in this way we find that the product of any 
number n of factors, each of the form 



cosa+V— 1 sin a = cos (a + ft + y -\ n terms) 

+ V— 1 sin (a + ft + y -\ n terms). 

Suppose now that a = ft = y = etc. = 6, then we have 
(cos0+ V— 1 sin0)*=cosn0+V^n sinnO, 
which proves the theorem when n is a positive integer. 

* From the name of the French geometer who discovered it. 



230 PLANE TRIGONOMETRY. 

II. When n is a negative integer. 

Let n — — m ; then m is a positive integer. Then 



(cos 6 + V— 1 sin 6) n = (cos + V^l sin 6)~ m 

1 _ 1 

(cos#+ V— 1 sin0) m cosm0+V— 1 sinm0 

1 cos mS - V— 1 sinm# 

x 



(by I.) 



cosm# + V— 1 sinm0 cosm# — V— 1 sinm# 

cosm# — V— 1 sinm# * / — r . * 

= — = cos rnv — V — 1 sin ra0 

cos 2 m# + snr mO 

= cos ( — m6) + V — 1 sin ( — m&) . 

.-. (cos#+V— 1 sin0) w = cosw0+V— 1 sin nO, 
which proves the theorem when n is a negative integer. 

III. Wlien n is a fraction, positive or negative. 

P 
Let n = -, where p and q are integers. Then 

(cos0+V—l sin^^cospfl+V^Tsinpfl (by Land II.). 
But (cos-tf+V^T sin-0J =cosp^+V :r lsin^. 

. \ (cos + V^ sin $y = ( cos -6+ V^T sin 1 Oj. 

.-. (cos<9 + V^ : lsine)* = cos-(9 + V :: nsin^(9; 

p 

Chat is, one of the values of (cos + V— 1 sin 9) q 

p0 , /— T . P0 
js cos h V — 1 sin — 

g 2 

In like manner, 

(cos 6 — V— 1 sin 0) n = cos n6 — V— 1 sin ?i0. 

Thus, De Moivre's Theorem is completely established. 
It shows that to raise the binomial cos# + V— lsin0 to 



BE MOIVRE'S THEOREM, 231 

any power, we have only to multiply the arc by the 
exponent of the power. This theorem is a fundamental 
one in Analytic Mathematics. 

154. To find All the Values of (cos 6 + V^l sin fl)l — 

When n is an integer, the expression (cos 6 + V— 1 sin#) n 

P 

can have only one value. But if n is a fraction = -, the 

expression becomes 



(cos 6 + V^T sin 6) * = \ (cos + V— 1 sin 0) p , 

which has q different values, fron* the principle of Algebra 

(Art. 235). In III. of Art. 153, we found one of the values 

p 

of (cos 8 + V— lsin#) 2 ; we shall now find an expression 

p 

which will give all the q values of (cos + V — 1 sin 6) q . 

Now both cos 6 and sin 6 remain unchanged when 6 is 

increased by any multiple of 2tt\ that is, the expression 

cos0 + V— lsin0 is unaltered if for we put (Q + 2rir), 

where r is an integer (Art. 36). 

... (cos 0+ Vising) *" 

= [cos (0 + 2ttt) + V^~I sin (0 + 2rir)f 

= cos^^ + 2r7r) +V :r l^ P(g+ / r7r) (Art.l53)(l) 

The second member of (1) has q different values, and no 
more; these q values are found by putting r=0, 1, 2, ••• g— 1, 
successively, by which we obtain the following series of 
angles. 

p(6+2rTr) pO 

IT 

p(6 + 27r) 



When r = 0, cos = cos - 



r = 1, " = cos 

r = 2, u = cos 

etc. etc. 



P(6+±7T) 



232 PLANE TRIGONOMETRY. 

When r=q-l, cos^ +2 ^ = ^^ + 2,^-1)] 

q q 

p(6 + 2qTT-2w) 

= cos -• 

1 

All these q values are different. 

p(0 + 2r*) p(8 + 2q*) 

When r = q, cos — = cos 

* q q 

fp6 „ \ pd 

= cos \-jj- + 2pTrj= cos —, 

the same value as when ¥= 0. 

p(6 + 2r7r) [0 + (2g + 2)7r] 

When r = q + l, cos = cos — — 

= cos^±M ? 
8 

the same as when r == 1, 

etc., 
from which it appears that there are g and only q different 

values of cos — — , since the same values afterwards 

Q 
recur in the same order. 

«..,,, • P(0 + 2r7r) 
Similarly for sin • 

Therefore the expression 

p(6 + 2rir) t . p(6 + 2T7r) 

cos — + V — 1 sin- - 



gives aZZ the q values of (cos 6 + V— 1 sin#) g and no more. 
And this agrees with the Theory of Equations that there 
must be q values of x, and no more, which satisfy the equa- 
tion x q = c, where c is either real or of the form a+6V-l. 






APPLICATIONS OF BE MOIVRE 1 S THEOREM. 233 
APPLICATIONS OF DE MOIVRE'S THEOREM. 

155. To develop cos nd and sin nO in Powers of sin and 

COS0. 

We shall generally in this chapter write i for V — 1 in 
accordance with the usual notation. 

By De Moivre's Theorem (Art. 153) we have 

cosn# + i sinnQ = (cos 0-\-ismO) n (1) 

Let n be a positive integer. Expand the second member 
of (1) by the binomial theorem, remembering that i 2 =— 1, 
i 3 = — i, and that i 4 = + l (Algebra, Art. 219). Equate 
the real and imaginary parts of the two members. Thus, 

cos nd = cos" - - ( n ~ - 1 ) cos"" 2 6 sin 2 6 

n (n- 1) (n - 2) (n - 3) ^.^ ■ 4g _ etc> (2) 

Li 

sin n0 = n cos n ~ l sin 6> - n ( n ~ ^ ( n ~ 2 ) cos"- 3 fl sin 3 <9 

I? 
n(n _l) (n _ 2 ) (n-3) (n - 4) cos n- 5 ^ sin 5g_ etc< (3) 

The last terms in the series for cos nO and for sin nO will 
be different according as n is even or odd. 

The last term in the expansion of (cos 8 + ism6) n is 
i"sin"0; and the last term but one is ra^cos 6 sin"- 1 ^. 
Therefore : 

When n is even, the last term of cosn# is ^"sin"0 or 

n 

( — 1) 2 sin n 6, and the last term of sin nO is ni n ~ 2 cos sin* -1 9 

n-2 

or n(— l)" 2 " cos sin" _1 0. 

When n is odd, the last term of cos nO is ni n ~ x cos sin" _1 

n-l 

or n(— 1) 2 cos sin" -1 0, and the last term of sinn0 is 

n—l 

i n - l sm n 0oi (— 1) 2 sin"0. 






234 PLANE TRIGONOMETRY. 

EXAMPLES. 

Prove the following statements : 

1. sin 40 = 4 cos 3 sin — 4 cos sin 3 0. 

2. cos 4 (9 = cos 4 (9-6 cos 2 sin 2 (9 + sin 4 (9. 

156. To develop sin and cos in Series of Powers of 0. 

Put n8 = a in (2) and (3) of Art. 155 ; and let n be in- 
creased without limit while a remains unchanged. Then 

since — -, must diminish without limit. Therefore the 
n 

above formulae may be written 
cosa = cos n ^ ^cos n 2 0' 



[ «(a-g)(a-2fl)(«-3g) coslt ^ g /sing\ (1) 



[2 V 

I* 

j • n-i /i /sin 
and sin a = a cos n_1 

V » 
_ «(« - fK« - 2fl) e0S »Wgingy + ... _ . (2) 

If n = oo ? then = 0, and the limit of cos and its powers 

is 1; also the limit of ( J and its powers is 1. Hence 

(1) and (2) become \ d ' 

cos a = 1 — — + — — — H — (3) 

a 3 .a 5 //i\ 

[3 [5 

Sch. In the series for sin a and cos a, just found, a is ^e 
circular measure of the angle considered. 



CONVERGENCE OF THE SERIES. 235 

Cor. 1. If a be an angle so small that a 2 and higher 
powers of a may be neglected when compared with unity, 
(3) becomes cos a = 1, and (4), sin a = a. 

If a 2 , a 3 be retained, but higher powers of a be neglected, 
(3) and (4) give 

sin a = a — —; cos a = 1 (Compare Art. 134) 

Cor. 2. By dividing (3) by (4), we obtain 

. , a 3 t 2 a 5 t 17 a 7 , , /trx 

157. Convergence of the Series. — The series (3) and (4) 
of Art. 156 may be proved to be convergent, as follows : 

The numerical value of the ratio of the successive pairs 
of consecutive terms in the series for sin a are 

a 2 a 2 a 2 a 2 , 
5 etc. 



2-3 4-5 6-7 8-9' 

Hence the ratio of the (n + l)th term to the nth term is 

a 2 
; and whatever be the value of a, we can take 

2n(2w + l)' 

n so large that for such value of n and all greater values, 
this fraction can be made less than any assignable quantity; 
hence the series is convergent. 

Similarly, it may be shown that the series for cos a is 
always convergent. 

158. Expansion of cos n in Terms of Cosines of Multi- 
ples of 6, when n is a Positive Integer. 

Let x = cos 6 + i sin 6 ; 

then - = = cos 8 — i sin 0. 

x cos + i sin 

.-. cc + - = 2cos0; and x = 2isin0 (1) 

xx v ' 



236 PLANE TBIGONOMETBY. 

Also X n =(cos6+ism6) n =cosn0+ismn6 (Art. 153) (2) 

and — = (cos — i sin 0) n = cos n9 — i sin nO . . . (3) 

x 

.\ 2 cos n6 = x n -\ , and 2 i sin nO = x n . (4) 

x n x n w 

Hence (2 cos 0) w = (x + x~ l ) n , by (1), 
= x n + nx n ~ 2 + n ( n — 1 ) x n ~* + etc. + m~ in " 2) + x~ n 

V x n J V x n ~ 2 J [2 v x n -y 

= 2cosn6 + n2 cos (n - 2)6 + n ( n ~ 1>} 2cos(n - 4)0 + etc. 

.\ 2 W_1 cos n 6 = cos w0 + n cos (w — 2) 

+ ^C^" 1 ) cos (n - 4) + etc. (5) 

[£ 

Note.— In the expansion of (a? + a; —1 ) w there are n + 1 terms; thus when n is 
even there is a middle term, the (- + ljth, which is independent of 0, and which is 

n(n-l)...(-+l) 
n(n-l)...(n-hn + l) . ___L_L 

\\n ' ' '* \±n 

Hence when n is even the last term in the expansion of 2 n ~ l cos n 6 is 
n(n-l)...(2+l) 

Si 

When w is odd the last term in the expansion of 2 W—1 cos n 9 is 
n(n-l)...£ (w + 3) ftna a 

159. Expansion of sin n in Terms of Cosines of Multiples 
of 0, when n is an Even Positive Integer. 

(2 i sin 0)" = fx - -Yby (1) of Art. 158 

= a* - nx n ~ 2 + n ( n "" ^ of" 4 + ... 

7l(n — l) - (n -4) _ ^-(n-2) . %-n 

|2 



EXPANSION OF SIN" 6. 



237 



. n ( X n-2_^) + H]}_=l) 



x n ~*+ 



+ 



(-l)"»(n-l)...7| + l 



•. 2"- 1 (— iy sin n ( 



= cos nd — n cos (n — 2) # + ^^ '- cos (w — 4) 6 



+ 



(-!)•«(»- 1) ...(2 + 1 



2^n 



160. Expansion of sin" in Terms of Sines of Multiples of 
6, when n is an Odd Positive Integer. 

(2 i sin e) n =fx- -Yby (1) of Art. 158 

= sb" - nx n ~ 2 + ™ ( n ~ - 1 ) a;"- 4 



12 



[2 



(n-2) ^-n 



a;" I x"-V |2 V . a?"- 4 



+ " li(n-l) l *) 

: (2 i sin 6) " 

= 2 i sin n0 — n 2 i sin (n — 2) 

+ w ("- 1 ) 2tsin(n-4)fl 



+ (-l)»n(n-l) ^( w + 3) 2 . sing [(4)ofArt . 158] 
ti (» — 1) 



238 PLANE TRIGONOMETRY. 

Whence dividing by 2i, we have 

w-l 

2 W ~ 1 (-1) 2 sin w 

= sin 710-rc sin (ti - 2)0 + n ( n " 1 ) gin (n - 4)0 + ... 

(-l)^n(n-l)...i(n + 3) 

IK*-*) 

EXAMPLES. 

Prove that 

1. 128 cos 8 0=cos 8 0+8 cos 60+28 cos 40+56 cos 20+35. 

2. 64 cos 7 = cos 70 + 7 cos 50 + 21 cos 30 + 35 cos 0. 

161. Exponential Values of Sine and Cosine. 

Since e* = 1 + ^ + ^ + ^ + ,- + — . . . (Art. 129) 

|2^[3 [4 v ' 

0^ s 
[2'|4 '"V |3 + [5" 
= cos + * sin , . (Art. 156) 

and e - = l-| + ^-etc.- t -^-| + | |-etc.) 

= cos — i sin 0. 
.-. 2cos6 = e i6 + e- id , and 2 1 sin = «*'* — e-* . . (1) 

g*0 _I_ g—iO giO p—%9 

.-. cos = — — , and sin = • . . . (2) 

2, ' 2i w 

which are called the exponential values* of the cosine and 
sine. 

Cor. From these exponential values we may deduce 
similar values for the other trigonometric functions. Thus, 

piO p-i9 , n 

tan0 = -^ —r~ (3) 

* Called also Euler's equations, after Euler, their discoverer. 



^-etc.) 



GREGORY'S SERIES. 239 

Sch. These results may be applied to prove any general 
formula in elementary Trigonometry, and are of great im- 
portance in the Higher Mathematics. 

EXAMPLES. 

-i T3 sin 2 , „ a 

1 Prove rr^^ =tanft 

We We 2i ^ 26 = ^~^\ by(1) 
2 + 2cos20 2 + e 2 '« + e-2»» J v ' 

piO p— iO 

■ itan0 by (3). .\ etc. 



e 10 + er 

Prove the following, by the exponential values of the 
sine and cosine. 

2. cos 2 a = cos 2 a — sin 2 a. 

3. sin0 = -sin(-0). 

4. cos3 = 4cos 3 0-3cos0. 

Rem. — If we omit the i from the exponential values of the sine, cosine, and tan- 
gent of 0, the results are called respectively the hyperbolic sine, cosine, and tangent 
of 0, and are written sinh 9, cosh 9, and tanh 9, respectively. Thus we have 
sinh 9 = — i sin 19, cosh 9 = cos i9, tanh 9 = — i tan i9. 

Hyperbolic functions are so called, because they have geometric relations with 
the equilateral hyperbola analogous to those between the circular functions and the 
circle. A consideration of hyperbolic functions is clearly beyond the limits of this 
treatise. 

For an excellent discussion of such functions, the student is referred to such 
works as Casey's Trigonometry, Hobson's Trigonometry, Lock's Higher Trigonom- 
etry, etc. 

162. Gregory's Series. — To expand in -powers of tan 

where lies betiveen — - and + -• 

2 ' 2 

By (3) of Art. 161, we have 

pi9 Q—i9 



i tan = 



pi9 _|_ e -i9 



l+t'tanfl == 2e* = 2 „ 
l-itan0 2e~ ie 



240 



PLANE TRIGONOMETRY. 



.\ log e 2i9 = log(l + i tan 0) - log(l - i tan 0). 

.-. 2 iff = 2 i(tan — \ tan 3 + £ tan 5 - etc.) (Art. 130 ) 

.-. = tan0 — £tan 3 0+i-tan 5 — etc (1) 

which is Gregory's Series. 

This series is convergent if tan = or < 1, i.e., if lies 

between — ~ and ~, or between |tt and fir. 
4 4 

/Sc/i. This series may also be obtained by reverting (5) 
in Cor. 2, Art. 156. 

Cor. 1. If tan = x, we have from (1) 

x° J , x 5 



tan -1 #= x ■ 



3 + ^- etC> 



(2) 



Cor. 2. If = -, we have from (1) 

J-1-i+i-i+ete. 



(3) 



a series which is very slowly convergent, so that a large 
number of terms would have to be taken to calculate ir to a 
close approximation. We shall therefore show how series, 
which are more rapidly convergent, may be obtained from 
Gregory's series. 



163. Euler's Series. 

tan -1 - + tan" 1 - = - 
2 3 4 



(by Ex. 2, Art. 60) 



Put 6 = tan" 1 1. 



gives 

4. -I 1 1 

tan 1 - = - ■ 
2 2 



tan0 = ^, which in (1) of Art. 162 



3-2 3 5-2 5 7-2 7 



+ etc (1) 



Put = tan" 1 -. 
3 

tan- x - = -- 

3 3 3-3 3 



tan = -, and (1) becomes 
o 



1 + a 



5-3* 7-3' 



+ etc. 



(2) 



MACHINES SERIES. 241 



Adding (1) and (2) we have 

•) (3) 



\ n 






4 V 2 3 * 23 5 * 25 / V 3 3 - 33 5 - 3 ' 

a series which converges much more rapidly than (3) of 
Art. 162. 

164. Machin's Series. 

Since 2 tan" 1 - = tan" 1 — i— (by Ex. 3, Art. 60)= tan" 1 —, 
5 1— A 12 

. •. 4 tan" 1 - = 2 tan" 1 A = tan" 1 4 = tan" 1 — . 

5 12 1- T 2 A 119 

1 90 12.0 _ i 1 

Also, tan" 1 — - tan" 1 1 = tan" 1 J-ls — - = tan" 1 — . 
119 1 + m 239 

.v 4 tan -1 = tan -1 

5 4 239 

.-. | = 4 tan- 1- tan- £. 
* = 4 fl 1 1 



5 3-5 3 5-5 5 
/ 1 1 






V239 3-(239) 3 5(239) 5 ^ 
In this way it is found that it = 3.141592653589793 •• 

Cor. Since tan" 1 — + tan" 1 — = tan" 1 ?V + 2T9 _ 
99 239 1-^x^ 

= tan -1 — , 

70' 



.-. - = 4 tan- 1 | - tan- 1 — + tan" 1 — • 
4 6 70 99 

Kote. — The series for tan" 1 — and tan" 1 — are much more convenient for pur- 

70 99 * 

poses of numerical calculation than the series for tan -1 — • 
* 239 

Example. — Find the numerical value of n to 6 figures by 
Machin's series. 



242 PLANE TRIGONOMETRY. 

165. Given sin = x sin (0 + <*); expand 6 in a Series of 
Ascending Powers of x. 

We have e ie — e~ ie = x [e id+ia — e -tt-*«] . . (Art. 161) 
... e 2io _ i — x [e lid • e ia — e-* a ] 

1 — xe ia 
.•. 2i0=slog(l — xer**) — log(l — a;e* a ) 

— a5( e <a _ e -ia) _)_ 2Lf e 2*a_. e -2<a) _j_ ^L( e 3ia__ e -3fa) ... 

2 3 (Art. L30) 

.-. (9 = sb sin a + ^ sin 2 a + ~ sin 3a+— (Art. 161) (1) 

Example. If a = ir — 20, then 8 = 1. .*. (1) becomes 
= sin 20 - i- sin 40 + | sin 6(9 - J sin 8 6 + •-. 

166. Given tan x = ?i tan 6 ; expand & in Powers of n. 

— - — n— _ .... (Art. 161) 

e 2ix __ I 

■ssa n 



e w — 


e 


{0 


e 2i0 _ 


- 1 


iO 



e 2ix + i e* i9 + l 

. C 2 fa= (g w + 1 ) + *(*»"-!) 
(e2^ + l)-w(e 2W -l) 

(l + w)e 2<g + l— n 
~ (l_n)e 2 *' + l + n 

c 2W +wi / i l — n\ 
= ^ f where m = ] 

1 + n J 



= e 



me 2i9 + l \ 
f\ + me - M 



1 + roe- 2 **Y 

1 + WW 2 ** / 

.-. 2te = 2/0 + log(l +me- 2 ^)-log(l + me 2 *) 

= 2/0 - m(e 2W - e~ M ) + — ( e 4W - e~* ie ) . 

.\ a==0~wsin20 + ™~sin40 . . (Art. 161) 



RESOLUTION INTO FACTORS. 243 

RESOLUTION OF EXPRESSIONS INTO FACTORS. 

167. Resolve x n — 1 into Factors. 

Since cos 2 r-rr ± V — 1 sin 2 ttt = 1, 

where r is any integer, and # n = 1, 

.-. # n = cos 2r-n- ± V— 1 sin 2r?r. 

i 

.-. aj = (cos2nr±V-lsin2?^r)" 






= cos — ± V^l sin — * . . (Art. 153) (1) 
n n 

(1) Wlien n is even. If r = 0, we obtain from (1) a real 

root 1 ; if r = - -, we obtain a real root — 1, and the two cor- 

JS 

responding factors are x — 1 and a? + 1. If we put 
r = l,2,3...|-l, 

in succession in (1), we obtain n — 2 additional roots, since 
each value of r gives two roots. 

The product of the two factors, which are 






( a? — cos V — 1 sin 

\ n n 

and f a — cos f- V — 1 sin __ 

V n n J 

= f a? — cos + sin 2 

\ n J n 

= z 2 -2acos— +1 (2) 

which is a real quadratic factor. 

.\ z n -l= (^—1)^—20 008 — +l)(a 2 - 2zcos — + lV- 
a; 2 -2^cos ? ^7r + lYa; 2 ^2a;cos^^7r+iy- (3) 

* This expression gives the n nth roots of unity. 



244 PLANE TRIGONOMETRY. 

(2) When n is odd. The only real root is 1, found by- 
putting r = in (1) ; the other n — 1 roots are found by 

Yl \ 

putting r = l, 2, 3, ••• — - — in (1) or (2) in succession. 
... af — la. (x-l)(x 2 -2xcos — + lYo 2 - 2a? cos — + lV- 

..A 2 -2acos?^7r+lY^^ (4) 

168. Resolve x n + l into Factors. 

Since cos (2r + 1)tt ± V^l sin (2r + 1)tt = — 1, 
where r is any integer, and x n = — 1, 

.\ aj n = cos(2r + l)7r±V^Isin(2r + l)7r. 

i 
.-. a? == [cos(2r + l)7r±V— lsin(2r + l)7r] n 

2r + l /— r 2 ^+l 

= cos 7r±V— Ism 7r . . . (1) 

n n x ' 

which is a root of the equation x n = — 1; i.e., — 1 is a root. 

(1) When n is even. There is no real root; the n roots 
are all imaginary, and are found by putting 

r = 0,l,2,...|-l, 

successively, in (1). 

The product of the two factors, 

y n n J 

and ^-cos'21+i^+V^lsin^B^ 

\ n n J 

= aj«-2a?cos^±^7r + l (2) 

n v ' 

which is a real quadratic factor. 



RESOLUTION INTO FACTORS. 245 

x n + Jrf _ 2x cos - + lY^ 2 - 2ajcos — + 1 N 



../a 2 -2£COS ? -^^7r + lY^^ (3) 

(2) When n is odd. The only real rot)t is — 1 ; the 

7t 3 

other n — 1 roots are found by putting r = 0, 1, 2, ••• — - — ■ 

in (1), in succession. 

.-. x n + l= (x + l)(a? - 2x cos £ + A(x 2 - 2a? cos ^ + lY •• 

../a 2 - 2x cos ^^-^tt + lYa 2 - 2x cos ^^^tt + l) (4) 

EXAMPLES. 

1. Find the roots of the equation x 5 — 1 = 0. 
Ans. 1, cos^(2r7r) + isin4-(2?'7r) ? where r = 1,2,3, ,4. 

2. Find the quadratic factors of # 8 — 1. 
Ans. (x 2 - 1) (x*--V2x + 1) (a 2 + 1) (a 2 + V2 a + 1). 

3. Find the roots of the equation x 4 + 1 = 0, and write 
down the quadratic factors of x* + 1. 

^h*. ±- L ^±V~ := 3~4:; (x 2 -W2 + l)(f + W2 + l). 
V2 V2 

169. Resolve xr n — 2 x n cos + 1 into Factors. 

Let x* n — 2a? cos (9 + 1 = 0. 
.-. ft 2 " — 2# n cos 8 + cos 2 6 = — sin 2 0. 

.-. a? n — cos 6 = ± V— 1 sin = ± i sin ft 

... x = (cos (9 ± i sin 0)» = cos 2r7r + ° ± i s in 2r?r + (9 (1) 

n n 

since cos 6 is unaltered if for we put + 2 r7r. If we put 
r = 0, 1, 2, •••w — 1, successively in (1), we find 2n differ- 
ent roots, since each value of r gives two roots. 






246 PLANE TRIGONOMETRY. 



The product of the two factors in (1) 

2 Tit + 6 . • 2rir + ff 
= { x — cos i sin — 

n n 

X[x — cos J \- i sin — 

V n n 



== x 2 — 2 # cos — + 1 (2) 



n 

x 2n __ 2 a?" cos (9 + 1 



= f^ 2 -2^cos- + lV^-2^cos 27r + g + lY 
n J\ n J 

...( x 2 - 2 a cos- J + 1 

(2n-2)7r + (9 \ 
a 2 -2 a cos- J + 1J— (3) 

(7o?\ Change a; into - in (3) and clear of fractions, and 
a 

^.2n . f ~a o v 



we get # 2n — 2 a n af cos + a 2n = ( ^ — 2 a# cos - + a 2 

.~(x 2 - 2 ax cos ^+^ + a 2 Y^ 2 ~2ax cos i^+i + A.. 
V n A n J 

• • • to n factors (4) 

EXAMPLES. 

Find the quadratic factors of the following : 

1. a 8 -2 a 4 cos 60° + 1 = 0. 

Ans. (x 2 - 2x cos 15° + l)(ar - 2 x cos 105° + 1) 

X (x 2 - 2 x cos 195° + 1) (x 2 - 2 x cos 285° + 1) = 0. 

2. a; 10 -2^cosl0 o + l=0. 

Ans. (x 2 - 2 x cos 2° + 1) (.t 2 -2a? cos 74° + 1) 
X (a 2 - 2 a? cos 146° + 1) (x 2 - 2 x cos 218° + 1) 
(ar — 2 a? cos 2U0° + 1) = 0. 



BE MOIVBE'S PROPERTY OF THE CIRCLE. 247 

»170. De Moivre's Property of the Circle. — Let be the 
centre of a circle, P any point in its 
plane. Divide the circumference into 
n equal parts BC, CD, DE, ..., begin- 

(ning at any point B; and join and 
P with the points of division B, C, 
D, ... Let POB = 0; then will 
OB 2n -20B" • OP* cos n6l+OP 2n 




= PB' - PC - PD*. • • to n terms. 

Tor, put OB = a, OP = x, and 6 = -; then 

n 

PB 2 = OP 2 + OB 2 - 2 OP • OB cos 9 

= x 2 + a 2 — 2ax cos - (1) 

n v 

PC 2 = OP 2 + OC 2 - 2 OP . OC cos a + 2ir 

n 

= x 2 + a 2 -2ax cos a + 2?r : and so on . . (2) 
n 

Multiplying (1), (2), (3),— together, we have 
PB 2 • PC 2 • PD 2 ... to n terms 

= (x? - 2ax cos - + a 2 Yar* -2ax cos a + 2ir + ^ 

x(tf-2axcos^+^+a 2 \.- 

= a? n -2 a n x n cos a + a 2n . [by (4) of Art. 169] 

= O^-2OP n .OB n cosn0 + OB 2n ... (3) 

which proves the proposition. 




248 PLANE TRIGONOMETRY. 

171, Cote's Properties of the Circle. — These are particu- 
lar cases of De Moivre's property of 
the circle. 

(1) Let OP, produced if necessary, 
meet the circle at A, and let 

AB = BC = CD, etc.,-*—; 

n 

then nO is a multiple of 2tt. Hence 

we have from (3) of Art. 170, after taking the square root 

of both members, 

OB* _ oP = PB • PC . PD ... to n factors ... I. 

(2) Let the arcs AB, BC, •••be bisected in the points 
a, b, ••• ; then we have, by (1), 

OB 2 * _ op 2 * = Pa • PB • P6 • PC • Pc ••• to 2n factors. 
Hence, by division, 

OB w + OF = Pa • Tb • Pc ... to n factors ... II. 

Cor. If the arcs AB, BC, ••• be trisected in the points 
<hj a 2> &u &2> • ■ •? then we have 
OB^ + OF , -OG^+OP* i ===Po 1 .Pa 2 .P6 1 -P6 2 --- to 2n factors. 

172. Resolve sin 6 into Factors. 

(1) Put x= 1 ; then we get from (3) of Art. 169 

2(l-cos0)=2»(l-cos^l-^ 

-Yl-eoe< ai> -p' + ^ (1) 

Put 6 = 2ji<f> in (1), and let 2na = ir. 

1 — cos = 1 — cos2?i<£ = 2sin 2 n<£; 
then extracting the square root, we have 

sin n<f> = 2 n_1 sin<£ . sin (<£ + 2a) sin (<£ + 4a) x 

... xsin(<£ + 2na — 2a) (2) 









RESOLUTION INTO FACTORS. 249 

But sin (cf) + 2na — 2a) ; = sin (</> + tt — 2a) = sin (2 a — <f>), 

sin (<£ + 2na — 4a) = sin (4 a — </>), and so on. 

Hence, when n is odd, multiplying together the second 
factor and the last, the third and the last but one, and so 
on, we have 

sin n<f> 

= 2 n-1 sin <£ sin (2 <* + <£) sin (2 a— <jf>) sin (4 «+</>) sin (4 a— <£) 
• ••X sin [(?*, — 1) a + </>] sin [(n — l)a — <£]. 

But sin (2a + <f>) sin (2 a— <£) = sin 2 2a — sin 2 </>, and so on. 

.\ sinn^>=2 n_1 sin^)(sin 2 2a — sin 2 <£) (sin 2 4a — sin 2 <£) x — 
... X [sin 2 (?i — l)a — sin 2 <£] .... (3) 

Divide both members of (3) by sin <£, and then diminish 
</> indefinitely. Since the limit of sin n<£ -s? sin <£ is n, we 
get 

n = 2 n ~ 1 sin 2 2a sin 2 4a sin 2 6cc x ••• X sin 2 (w — l)a (4) 

Divide (3) by (4) ; thus 

sin^ = ,sin^l-^Yl--^) X ... (5) 
\ sm 2 2a/\ sin 2 4 a/ 

Put ncf) =s 0, and let n be increased while <£ is diminished 
without limit, 6 remaining unchanged ; then since 2 na = ir, 
the limit of 

. sm 2 - sm 2 - [ - ] 
sm 2 (j) 7i _ <9 2 n W __ (9 2 

sm 2 - 

id the limit of n sin <£ = that of ti sin - = 0; and so on. 
Hence (5) becomes 

*r( l -3( i -&X i -&}" • <6) 

Note. — The same result will be obtained if we suppose n even. 



250 PLANE TRIGONOMETRY. 



ber 

1 is 



Rem. — When 0>O and <n, sin0 is +, and every factor in the second member 
of (6) is positive; when 0>7r and <2 7r, sin is — , and only the second factor 
negative; when 0>2 -n and <3 7r, both members are positive, since only the second 
and third factors are negative; and so on. Hence the + sign was taken in extract- 
ing the square root of (1). 

Cor. Let # = 7", then sin - = 1, and - = -• Hence (6) be- 

2 2 7r 2 

comes 



2 2 y V 2 2 - 2y \ 3 2 • 2 : 

7T 1^3 3jJ> 5-7 . 

2 ' 2 2 "' 4 2 ' 6 2 ~" 



7T 2 2 4 2 6 2 



2 1-3 3-5 5.7 7.9 
which is Wallis's expression for w. 

173. Resolve cos0 into Factors. — In (2) of Art. 172, 

change <f> into <£ + a, then n$ becomes ncj> + na, i.e., ncf> + -• 
Hence (2) becomes 

cos n<j> = 2 n ~ l sin(<£ + a) sin(<£ -f*3a) sin(<£ + 5a) 

X ...sin[> + (2n-l)«] (1) 

But sin(<£ + 2na — a) = sin(<£ + 7r — a) = sin (a — </>), 

sin(<£ + 2w« — 3a) = sin(3« — </>), and so on. 
Hence when n is even we have from (1) 
cosn<£ = 2 n_1 sin(« + <f>) sin (a — cj>) sin(3a + <£)sin(3ce — <£) 
X ••• X sin[(n — l)a + <£] sin[(?i — l)et — <£] 
= 2 w ~ 1 (sin 2 a - sin 2 <£) (sin 2 3« - sin 2 <£) 

X ••• X [sin 2 (n — l)r>c — sin 2 <£] (2) 

Therefore, putting n<j> = 0, as in Art. 172, we obtain 

Note. — For an alternative proof of the propositions of Arts. 172 and 173, 
Lock's Higher Trigonometry, pp. 92-95. 






SUMMATION OF SERIES. 



251 



EXAMPLES. 

1. If a = — , prove that 

4?i 

sin a sin 5 a sin 9 a • • • sin (4 n — 3) a = 2~ n+ *. 

2. Show that 

16cos(9cos(72°-^)cos(72 + ^)cos(144 -^)cos(144 + ^) 

= cos 5 6. 



SUMMATION OF TRIGONOMETRIC SERIES. 

174. Sum the Series 

sin a + sin(a + /?) + sin(« + 2(3) H h sin [a + (n — 1)/J]. 

We have 

2sin«sin|-/? = cos( « — £ j — cosf a + 2 j (Art. 45) 

2 sin (a + f3) sin|-/? = cos fa + | j — cos (a + -§-0), 

2 sin (a + 2/2) sin^/3 = cos (a + f /3) - cos (a + | /?), 
etc. = etc. 
2sin[> + (ft--l)/3]siniL/? 

, 271-1 



COS 



. 2?i — 3 ~ 
a ^ o — /* 



— cos 



:/? 



Therefore, if S B denote the sum of » terms, we have, by 
addition, 



2S„sin£/J = cos (« — 1-/3) — cos 
, n-1 



= 2 sin 



sin 



S. = - 



« + 



2 
w-1 



: /3 



, 2n-l / ,~| 
sin|n/3 . . (Art. 45) 







sin |- ri/3 



sin£/3 



252 



PLANE TRIGONOMETRY. 



175. Sum the Series 

cos a + cos (a + /}) + cos (« + 2/3) H + cos [a+ (?i — 1)/3]. 

"We have 2 cos a sin ■$■ /J = sin (a + |-/3) — sin (a — i/3), 
2 cos (a + 0) sin££ == sin (« + f £) - sin (« + i/3), 
etc. = etc. 
2 cos [a + (n - l)/3] sin J/3 
, 2ti-1 



: sin 



: /? 



— sin 



. 271-3. 



'} 



Denoting the sum of n terms by S n , and adding, we get 

2n-l 



2S n sin ^-/3 = sin 



cos 



S„. = ■ 



a + - 



a + 



: P 



-/3 -sin (a -J/3), 
sin Jw/3 



sin££ 

i?em. — The sum of the series in this article may be deduced from that in Art. 174 
by putting a + - for a. The sums of these two series are often useful;* and the 
student is advised to commit them to memory. 

— 7T 

Cor. If we put /3 = — , then sin Jn/3 = sin ?r = 0. Hence 
we have from Arts. 174 and 175 

■(■ 

cos tt + cosf a ~^~ * j + cos( a-\ — - ) + • 



sin«+sin[ a+ :L -^ ) + sin( a + ^- ^ sin 



2j 
n 



4?r 
n 



2(n-l). 



r ,2(7»-i) i 

a-\ — s '-ir 

L » j 



= 0. 



Note. — These two results are very important, and the student should carefully 
notice them. 

176. Sum the Series 

sin m a+sin m (a + /3)+sin m (a+2/3)H |-sin m [a+(n — 1)/3]. 

This may be done by the aid of Art. 159 or Art. 160. 

* See Thompson's Dynamo-Electric Machinery. 3d ed., pp. 345, 346. 



SUMMATION OF SERIES. 253 

Thus, if m is even, we have from Art. 159 

m 

2 7n ~ 1 sin m a = ■( — l) 2 [cos ma — m cos (m — 2)a+ •••] • (1) 

m 

= (-1)2 [cos m (<*+/?) -m cos (m-2) (<*+/?) +•••] (2) 

and so on ; and the required sum may be obtained from the 
known sum of the series 

[cos ma + cos m (a + /?) + cos m (a + 2 /J) H ] 

and jcos (m — 2) a + cos[(m — 2) (« + /?)] 

+ cos [(m-2) (a + 2 /})] + ...}, etc. 

We may find the sum of the series 

cos m a + cos m (a + /3) + cos w (a + 2 /J) + etc. 
to n terms in a similar manner by the aid of Art. 158. 

EXAMPLES. 

1. Sum to n terms the series 

sin 2 a + sin 2 (a + (3) + sin 2 (a + 2 /?) + ... 

We have 

2 sin 2 a = — (cos 2 a — 1) by (1), 

2 sin 2 (a + p) = - [cos 2 (a + /?) - 1] by (2), 

2 sin 2 (a + 2 /?) = — [cos 2 (a + 2 fi) - 1], and so on. 

Hence 

2 S n =?i — [cos 2 cc + cos 2(<x + /3) + cos 2(<*+2/3) +•••] 

san _ cos[2« + (n--l)ff]sinnff (A 175) 

sin/? ' V " • 

S — n cos [2 a + (n — l)/3]sinn/? f 
n ~2 2sin/3 



254 



PLANE TRIGONOMETRY. 



2. Sum to n terms the series 

cos 3 a + cos 3 2 a + cos 3 3 a + 



Ans 2 cos [3 a + \ Q - 1) 3 a] sin f na 



8 sin | a 



6 cos 



a + 



n — 



. na 
— =! sin — 



8 sin -i- a 
177. Sum the Series 

sin a — sin (a + (3) + sin (cc + 2 /?) — • • • to n terms 

Change (3 into /? + tt, and (1) becomes 

since + sin (« + 7r + /?) + sin(cc + 27r + 2 i 8)H . 

Therefore we have from Art. 174 



sin 



S„ = • 



(n-l)Qr + )8)1 nQ + /?) 

2 2 



sin — i-t- 
2 

Similarly, 

cos a — cos (a + /?) + cos (a + 2 /3) — • • • to n terms 



cos 



« + 



(«-l)( ff + j8 )- 



sin 



2 



sin- 



■ + /B 



(1) 
(2) 

(3) 



(4) 



178. Sum the Series 

cosec + cosec 2 + cosec 40 + 



to 7i terms. 



We have 



Q 

cosec = cot cot 0, 

2 ' 



cosec 2 = cot — cot 20, 
etc. = etc. 
cosec 2 n ~ 1 = cot 2 n - 2 - cot 2 2n ~ 1 0. 
Therefore, by addition, as in Art. 174, 



S n 



;C oti0-cot2 2n - 1 0. 



SUMMATION OF SERIES. 255 

Note. — The artifice employed in this Art., of resolving each term into the dif- 
ference of two others, is extensively used in the summation of series. 

Practice alone will give the student readiness in effecting such transformations. 
If he cannot discover the mode of resolution in any example, he will often easily 
recognize it when he sees the result of summation. 

The student, however, is advised to resort to this method of solution only as a last 
resource. 

179. Sum the Series 


l 2 ' * "4 



tan0 + -|-tan r + J tan T + ••• to n terms. 



We have tan 6 — cot — 2 cot 2 6, 

a n 

% tan - = -|- cot cot 6, 

z z 

itan| = icot^-icot| 

etc = etc. 

1.0 1 . 1 . 6 

■ — tan — = cot ■ cot — • 

on-i 2 n ~ l 2 n_1 2 n ~ 1 ' 2"~ 2 2 n ~ 2 

.: S„ = — cot— -2cot20. 

2 n 2 n 

180. Sum the Series 

sina+ajsin(a+j3)+as 2 sin(a+2/?)+ •••aj n ~ 1 sin[oc+(n— 1)/?]. 

Denote the sum by S n , and substitute for the sines their 
exponential values (Art. 161). Thus, 

2iS n = (e ia — e~ ia ) + aj(e*< a +P) — e~ *(* + £)) 

ei a _ ^ M r u _ x n e -n* + m [A1 ^ (3) ^ 163] 



1 — xe* 1 



xe~ 



^ e ia — e -ia — ^ ^gi(a-fl) _ g-f (a-ff) ] _a»n[V(a+W0) _g-t(a+W0) J 

1 — ^(c^ + c-^ + aj 2 

+ ft 71 * 1 [e* (tt+W|8-]8) — e -i(a+nf}-p) ] 

1 — x(e^ + e-^) + x 2 



256 PLANE TRIGONOMETRY. 

.'• S w = 

sing— xsm(a—/3)—x n sin(a+n^) + x n+1 sin [g+(n— -l)/3] ,-,n 
l-2o;cos/3 + o; 2 " ^ ; 

Cor. If # < 1, and n be indefinitely increased, 

S _ sin a — x sin (a — /3) ,^ 

°°"~~ 1-20JCOS/3 + OT 2 " ( ) 

Sch. Similarly, 
cos a + x cos (a + /3) + a; 2 cos (a + 2/3)+ ••• to n terms = 
cosa— #cos(a— /3)— o? n cos(a+?i/3) + o; n+1 cos[a+(n— l)/3] ,»>. 

l-2o?COS/3 + 0; 2 ^ 

We may obtain (3) from (1) by changing a to a + ^» 

Also ^cosg-gcoBQt-fl) 4) 

l-2acos/3 + o; 2 v J 

181. Sum the Infinite Series 

07sin(a + i 8) + |sin(a + 2 ) 8) + ^sin( a + 3 i 8)+..., 

If L? 

and a cos (a + /3) + ^ cos (a + 2 £) + ^ cos (a + 3/3) + .... 

[2 [3 

Let S denote the former series, and C the latter. 

Then C + i'S = ae*>+e) + -V>+ 2 0) + -e*(«+ 3 0) + ... 

[2 [3 

==e*/#e^ + -e* 2 £ + -e* 3 + ... ) 

= e ia( e ^ __ i) . . [ by ( 3 ) f Art. 129] 

= e**( e *(cos£ + £sm/3)_.:l) m m m (Art 161) 
— gc cos j3gi(a + x sin j3) gia 

= e* cos £[cos(a+#sin/3)+tsin(a+#sin/3)] 
— (cos a + i sin a) . . . . (Art. 161) 



EXAMPLES. 



257 






Equating real and imaginary parts, we have 
C = e xcoa P cos (a + x sin j3) — cos a, 
S = e XC0B P sin (a + x sin /?) — sin a. 



EXAMPLES. 

Prove the following statements : 

1. The two values of (cos 4 6 + V— 1 sin 4 0)^ are 

± (cos 2 (9 + V^^l sin 2 6) ... (Art. 154) 

2. The three values of (cos 6 + V— 1 sin 6)i are 

cos - + V — 1 sin -, cos — r [ — + v — Ism -I— , 

o o o o 

cos — -! h V — 1 sin J— . 



3. The three values of ( — l) 3 are 



1+V^3 



, -1, 



(Art. 154) 



4. The six values of (— l) 6 are contained in 

cos (2 r + 1)* ± V=l sin < 2r + 1),r > where r = 0, 1, or 2. 
6 6 

5. The three values of (1 + V— 1) ¥ are contained in 

2* 



(7 . / T • (7 

cos - + V — 1 sin - 
3 3 



, where = \ f tt, or - x ¥ 7 - tt. 



6. The three values of (3 + 4V— 1) 3 are contained in 



-\/5| cos 



2r7r + 



1 sin 



2 rTr + (9' 



, where r == 0, 1, or 2. 






[... , . . — 3 

7. cos 6 (9 = cos 6 (9-15 cos 4 <9 sin 2 (9 + 15 cos 2 sin 4 (9 - sin 6 0. 

8. sin 9 (9 = 9 cos 8 6 sin 6- 84 cos 6 sin 3 6 + 126 cos 4 <9 sin 5 

- 36 cos 2 (9 sin 7 (9 + sin 9 (9. 



258 PLANE TBIGONOMETEY. 

9. tan w0 

»taa«- »<»- 1 >(»-. 2 > fa»tf+... 

n 

"!_ "("-!) tan* e + n{n-l)(n-2)(n-3) 4 / 

10. Given ^2— = — — : show that is nearly the circular 

2166 J 

measure of 3°. 

Prove the following : 

11. - 64 sin 7 = sin 7 - 7 sin 5 + 21 sin 3 - 35 sin (9. 

12. -2 9 sin lo 0=coslO 0-lOcos 8 (9+45 cos 66>-120 cos4£ 

+ 210 cos 2 - 126. 

13. 2 6 (cos 8 + sin 8 0) = cos 8 (9 + 28 cos 4 + 35. 

14. cos 6 + sin 6 = 1(5 + 3 cos 40). 

15. Expand (sin0) 4n + 2 in terms of cosines of multiples 
of 0. 

16. Expand (sin 0) 4n+1 in terms of sines of multiples of 0. 

17. Expand (cos 0) 2n in terms of cosines of multiples of 0. 

Use the exponential values of the sine and cosine to 
prove the following : 

-,o sin0 ,0 

18. = cot — 

1 - cos 2 

19. If log (# + y V— 1) = a + /3 V— 1, prove that 

x 2 + y 2 = e 2a , and y = x tan /?. 



20. If sin (a + ft V — 1) = x + y V — 1, prove that 
x 2 cosec 2 a — y 2 sec 2 a = 1. 



21. 2cos(ncos- 1 a?) = (^ + V :r rVl-^) r 



+ (z_V-l Vl-ar 2 )*. 



EXAMPLES. 259 






22. (^/-^iy- l = e 2 . 

23. e*(cos 6 + V^ 7 ! sin 0) = eV^cos j + V"ZT sin 

24. The coefficients of x n in the expansion, (1) of e ax cos bx, 
and (2) of e ax sin bx, in powers of x, are 

n w 

( a 2 4- ft 2 ) 2 / a 2 I ^2\ 2 

J — p- — *- cos n9 and ■* — j- — — sin w0. 

25. The coefficient of # n in the expansion of e x cos a? in 

n 

powers oi x is — cos 

* [n 4 

26. If the sides of a right triangle are 49 and 51, then 
the angles opposite them are 43° 51' 15"- and 46° 8 f 45" 
nearly. 

27. If a and b be the sides of a triangle, A and B the 
opposite angles, then will log b — log a 

= cos 2A — cos2B + |(cos4A — cos 4B) 

+ i(cos 6 A — cos 6B) -\ . 

28. If A + iB = log(m + m), then 

tan B = — , and 2 A = log(n 2 + m 2 ). 
m 

29. cos((9 + if) = cos o ( e ~* + ^ + i smef^^-^X 

30. sin ((9 + icfy) = sine( e ~^ + ^ - i coso( e ~ 



2 ) \ 2 

31. 2cos(« + i/3) = coscc(e 3 + e~P) — i sin «(e0 — e~0). 

32. (a + i6)( a +^) 

= r a e~^ r [cos(j81ogr + or) + z sin (/? log ?■ + ar)], 
where a + i6 = r(cos r + » sin r). 

33. log(a + ib) = 4 log(a 2 + 6 2 ) + * tan" 1 -. 

a 



260 PLANE TRIGONOMETRY. 

34. [sin(a — 0) + e± ia sin <9] n 

= sin" -1 ^ [sin(a — nO) + e ±ia sin n6~\. 



1-3 5 • 79- 11 

36. Write down the quadratic factors of # 13 — 1. 

Ans. (x — 1) [x 2 — 2 x cos y 1 ^ (2 r?r) + 1], six f actors, 
putting r = 1, 2, 3, 4 ; 5, 6. 

37. Solve the equation x 6 — 1 = 0. 

^Ins. (^ 2 -l)(^ 2 -a: + l)(^ 2 + ^ + l) = 0. 

38. Give the general quadratic factor of x?° — a 20 . 

Ans. x 2 — 2 ax cos -^ (V?r) + a 2 . 

39. Find all the values of a/I. 

-4ns. cos ^(ttt) + isin±(r7r), r having each integral 
value from to 11. 

40. Write down the quadratic factors of x e + 1. 

Ans. (x i -^/3x + l)(x 2 + l)(x i +V3x + l). 

41. Write down the general quadratic factor of x 20 + 1. 

Ans. x 2 — 2x cos (1 + 2 r) 9° + 1. 

42. Find the factors of x 13 + 1 = 0. 

Ans. (x + 1) [x 2 — 2 x cos T \ T (tt + 2 ttt) + 1], seven fac- 
tors in all. 

43. Find a general expression for all the values of V — 1. 

Ans. cos — ■ 1- i sin — , where r may have 

n n 

any integral value. 

44. Solve x 12 - 2 x e cos § tt + 1 = 0. 

.4ns. a? 2 — 2 # cos ^ (3 rw + ?r) + 1 = 0, six quadratics. 



EXAMPLES. 261 

45. Solve x 10 + V3 x 5 + 1 = 0. 

Ans. x 2 + 2 x cos (r x 72° + 6°) + 1 = 0, five quadratics. 

46. Write down the quadratic factors of 

x 2n — 2 x n y n cos a + y 2n . 

Aiis. x 2 — 2 xy cos — i — - -f. 2/ 2 ? ^ factors. 
Prove the following : 

47. tan <£ tan^ + -^tan/^ + ^V. tan^ + "ZllA- 

(-1)2, where n is even. [Use (2) of Art. 172.] 

48. sin 5 6 - cos 5 6 = 16 cos (0 - 27°) cos (0 + 9°) x 

X sin (6> + 27°) sin (0 - 9°) (cos - sin 0), 

49. *s + ,-* = 2fl+M ! Yl+ 4< 



50. € 0_e-* = 20(l + ^Yl + ' 



* -|_2 "^ 2 2 3 2 4 2 



l 2 3 2 5 2 T 



52. i + 4 + ^ + ^ + - = ' r ' 



53 o 36 144 324 576 
"" " 35 * 143 ' 323 ' 575 



51 


7T 2 


• 2 


• 4- 


4-6- 


6-8 


• 8... 




2 1 


• 3 


• 3- 


5-5- 


7-7- 


• 9 — 


55 


V2- 


4- 


36 


• 100. 


196- 


324 — 






3- 


35 


• 99. 


.195- 


323 — 


5f> 


Wfl: 


8 


; • 80 ■ 224 


= •440 


... 



9 • 81 • 225 • 441 



262 PLANE TRIGONOMETBY. 



57. cos x + tan ^ sin x = 

i+^Y*-— Yi+^Yi-— Yi+ 2 * 



58. cos a; — cot ^ sin x = 

2 



27r—yJ\ 2iT+yJ\ 4:7r—yJ\ ^-rr+y 

59. By aid of the formula cos = and Art. 172, 

J 2 sin <9 ' 

deduce the value for cos 6 obtained in Art. 173. 

60. By expanding both sides of Ex. 57 in powers of x 
and equating the coefficients of x, prove that 

to«=-?---JL + _2 ^-2_ + ^ ?_ + ... 

2 7T — ?/ 7T + 2/ 37T — 2/ 37T + 2/ 5 7T — ^ 5^ + ?/ 

61. Prove in like manner from Ex. 58 that 

,y 2 2 , 2 2 2 

2 ?/ 2rr — y 2ir + y 4?r — 2/ 4tt + 2/ 

^O T) ^ -, 1,1 1,1 1,1 

62. Prove F = l 

3V3 2 4 5 7 8 10 

a» -d ^ i 1,1 1,1 1,1 

63 - Pr ° Ve 2^ = 1 -5 + 7-n + 13-i7 + 19-- 

64. Prove that 



sin ?/ 



1 * |. y l_ T A-- s A-+ 



y ir—y 2-K—y n+y 2ir+y 3tt— y 4:ir-y 3tt+2/ 

Sum the following series to n terms : 

. 7i + l • net 

sin — ! — a sin — 

2 2 

65. sin a + sin 2 a+ sin 3 aH \- sin na = • 

• a 
sin- 



EXAMPLES. 263 

n + 1 * na 

cos — ■ — asm — 

2 2 

66. cos a-f -cos 2 a + cos 3 a-\ |-cos na = 



sin- 

2 



67. sm a + sin 3a + sino«H = — : ■• 

sin a 

sin2na 



68. cos a + cos 3a + cos 5a -\ = 



2 sin a 

69. sin»«+Bin'2«+siii'3«+-= n8m< '- sl ° ngC08(w+1)a - 

2 sin a 

rA 9 , 9 . 20 i wsina + cos(n + l)asin7ia 

70. cos-a + cos-2a+cos^3aH = ! * — ! — L 

2 -sin a 

71. sin 3 a + sin 3 (a + 13) + sin 3 (a + 2/3) H 

^m(a + --^-f3\sm-^ ^ml 3a + -^— 3/?Jsm-^ 
~~4 sin-i-/? 4 sinf/3 

72. sin 3 a + sin 3 2a + sin 3 3a H 

• na ■ /n + 1\ ■ 3 net . 3(n + l)a 
sm — sm f — ^— a sm sm — ^ — — — '— 

3 2 y 2 J 2 2 

4 • a A . 3a 
sm - 4 sm — 

2 2 

73. sin a sin 2a + sin 2 a sin 3 a + sin 3 a sin 4 a H 

_ n sin a cos a — sin na cos (n + 2)a 
2 sin a 

74. tana + 2tan2a + 2 2 tan2 2 aH = cot a — 2 n cot 2 n a. 

75. (tan a + cot a) + (tan 2 a + cot 2 a) + (tan 2 2 a + cot 2 2 a) 

+ ... = 2 cot a — 2 cot 2 W a. 

76. sec a sec 2 a + sec 2 a sec 3 a + • • • 

= cosec a [tan (n + 1) a — tan a]. 



264 PLANE TRIGONOMETRY. 

77. cosec a cosec 2 a + cosec 2 a cosec 3 a -\ 

= cosec a [cot a — cot (n + 1) a]. 

78 sin 2 sin 4 _ sec(2n + l)0— sec 

cos cos 3 cos 3 cos 5 2 sin 

79. cos 4 a + cos 4 (a + /?) + cos 4 (a + 2 0) H = f n + 

cos[2a+(n — l)/3]sinn/3 cos [4 a + (n — 1)2)8] sin2nft 

2 sin /J 8 sin 2 /? 

OA , ^ sin + sin 30 + sin 50 + ••• to n terms 

80. tann0 = — — — J — — ] - 

cos + cos 3 0'+ cos 50+ ---to n terms 

81. cos cos (0 + a) + cos (0 + a) cos (0 + 2 a) 

+ cos(0 + 2 ot)cos(0 + 3 a) + ••• 

n , cos (2 + net) sin not 

= - cos ot H - ' L 

2 2 sin ot 

82 s * n 0— sin2 + sin 3 to n terms , n + 1 , ^ 

cos0— cos2 0+cos3 to n terms 2 

83. sin(> + 1)0 cos + sin(# + 2)0 cos 2 0+... 

_ n sinp0 sin ( p + 1 + n) sin n0 

" 2 2sin0 

84. sin 3 sin + sin 6 sin 2 + sin 12 sin 4 + ... 

= i(cos20-cos2 n+1 0). 

85. sin /sin -Y+ 2. sin - f sin -Y+ 4 sin - f sin -Y+ - • 

V 2)^ A V 4 8/ 

= 2 n - 2 sin-^--isin2 0. 

9n-l 4 


86. tan - sec + tan - sec - + tan- sec — | — =tan 0— tan — • 

2 4 2 8 4 2 n 

87. cot cosec + 2 cot 2 cosec 2 + 2 2 cot 2 2 cosec 2* + • • > 

1 2"- 1 



2sin= 2 



2 sin 2 2"" 1 



+ — 



EXAMPLES. 
1 1 



265 



sin sin 2 sin 2 sin 3 sin 3 sin 4 

1 



+ 



89. 



sin0 
1 



(cot 3 — cot 4 0). 



sin cos 2 cos 2 sin 3 sin 3 cos 4 



:cosec(0 + - 



tan (« + 1) f + ! j - tai/fl + ! 



90. tan- 1 - 



1 + 1 + l 2 



tan - 



1 + 2 + 2 2 



-I- tan - 



1 + 3 + 3 2 



H = - — tan -1 • 

4 n + 1 



91. tan -1 a; + tan -1 



1 + 1 • 2 • £C 2 



•tan" 



a; 



1 + 2-3 -a? 

= tan -1 jkbl 



92. since sin 3« + sin-sm hsm — sin h - 

^ 2 2 2 2 2 2 



93. 



= - f cos -^— — cos 4 a V 

2 V 2 W - 2 



>+' 



:CC ]• 



cos + cos 3 cos + cos 5 cos + cos 7 

= ±-cosec0[tan(n + 1)0 — tan 0]. 

94. | sec + ^ sec sec 2 + — sec sec 2 sec 2 2 + . . . 

z z z 

= sin0(cot0-cot2 n 0). 

95. |logtan20 + ^logtan2 2 + - 3 logtan2 3 + ... 

Z Z Z' 

= log 2 sin 2 - - log 2 sin 2 W+1 0. 

2 n 

Sum the following series to infinity : 

% n . cos 0/1 . cos 2 Q/ , , cos 2 , ^ , 
. cos0H — cos 20 H — ■ cos 30 H — ■ cos 40+ ... 

\~L ■ 

= eco 8 2 cos ^ + gin q cos g^ 



266 
97. sin0- 



PLANE TRIGONOMETRY. 

sin20 , sin30 



12 13 



e -cos0 sin (sin 0). 



98. 1 _ cos20 + cos40 = i. cos ( cos ^ ^e + e -dn*y 

99. 2 cos (9 + f cos 2 + f cos 3 + f cos 4 + ... 



cos 



100. sin cos 0-} 



1 — COS0 
sin 20 cos 2 . sin 30 cos 3 



log(l — COS0). 



[2 



[3 

= e cos2 ^ sin (sin cos 0). 



101. cos0 + 5^cos20 + ^^cos30+... 

= e 8in e C08 cos (0 + sin 2 0) . 

102. sm0 + ^. S m20 + 5^sm30 + 

i Lz 

= e sm0cosd s i n (^ + sin 2 0). 



103. cos - ^cos 20 + icos 30 = log /2 eos-Y 

104. cos 20 + i cos 60 + i cos 100 + ... = ^log cot 







in* • /> a? 2 sin 20 , a 3 sin 30 .-./cosecfl , , A 

105. #sm0 = cot V hcot0). 

2 3 V x J 

106. x cos - - cos 20 + -cos 30 - -cos 40 + ... 

2 3 4 



107. sin0^-sin20?H^ + s i I1 3 ( 9 

1 2 3 



log(l + 2acos0 + o 2 ). 

sin 3 



— n.fti-1 



cot" 1 (l + cot 2 + cot0). 



108. 1 + 1 + 1 + 1+...=^. 
1^2 4 3 4 4 4 90 



109. i + i + i +! + ... = * 

1 4 ^3 4 5 4 7 4 



4 

96" 



PART II. 
SPHERICAL TRIGONOMETRY. 

CHAPTER X. 

POKMULE EELATIVE TO SPHEEIOAL TEIANGLES. 

182. Spherical Trigonometry has for its object the solu- 
tion of spherical triangles. 

A spherical triangle is the figure formed by joining any 
three points on the surface of a sphere by arcs of great 
circles. The three points are called the vertices of the 
triangle ; the three arcs are called the sides of the triangle. 

Any two points on the surface of a sphere can be joined 
by two distinct arcs, which together make up a great 
circle passing through the points. Hence, when the points 
are not diametrically opposite, these arcs are unequal, one 
of them being less, the other greater, than 180°. It is not 
necessary to consider triangles in which a side is greater 
than 180°, since we may always replace such a side by the 
remaining arc of the great circle to which it belongs. 

183. Geometric Principles. — It is shown in geometry 
(Art. 702), that if the vertex of a triedral angle is made 
the centre of a sphere, then the planes which form the 
triedral angle will cut the surface of the sphere in three 
arcs of great circles, forming a spherical triangle. 

Thus, let be the vertex of a triedral angle, and AOB, 
BOC, COA its face-angles. We may construct a sphere 
with its centre at 0, and with any radius OA. Let AB, 

267 



268 SPHERICAL TRIGONOMETRY. 







BC, CA be the arcs of great circles in which the planes of 

the face-angles AOB, BOC, COA 

cut the surface of this sphere; 

then ABC is a spherical triangle, 

and the arcs AB, BC, CA are its 

sides. 

Now it is shown in geometry 
that the three face-angles AOB, 
BOC, COA are measured by the sides AB, BC, CA, re- 
spectively, of the spherical triangle, and that the diedral 
angles OA, OB, OC are equal to the angles A, B, C, respect- 
ively, of the spherical triangle ABC, and also that a diedral 
angle is measured by its plane angle. 

There is then a correspondence between the triedral 
angle O-ABC and the spherical triangle ABC : the six 
parts of the triedral angle are represented by the corre- 
sponding six parts of the spherical triangle, and all the 
relations among the parts of the former are the same as 
the relations among the corresponding parts of the latter. 

184. Fundamental Definitions and Properties. — The fol- 
lowing definitions and properties are from Geometry, Book 
VIII. : 

In every spherical triangle 

Each side is less than the sum of the other two. 

The sum of the three sides lies between 0° and 360°. 

The sum of the three angles lies between 180° and 540°. 

Each angle is greater than the difference between 180° 
and the sum of the other two. 

If two sides are equal, the angles opposite them are 
equal ; and conversely. 

If two sides are unequal, the greater side lies opposite 
the greater angle ; and conversely. 

The perpendicular from the vertex to the base of an 
isosceles triangle bisects both the vertical angle and the 
base. 



DEFINITIONS AND PROPERTIES. 269 

The axis of a circle is the diameter of the sphere perpen- 
dicular to the plane of the circle. The poles of a circle are 
the two points in which its axis meets the surface of the 
sphere. 

One spherical triangle is called the polar triangle of a 
second spherical triangle when the sides of the first triangle 
have their poles at the vertices of the second. 

If the first of two spherical triangles is the polar triangle 
of the second, then the second is the polar triangle of the 
first. 

Two such triangles are said to be polar with respect to 
each other. Thus : 

If A'B'C is the polar triangle of 
ABC, then ABC is the polar triangle 
of A'B'C. 

In two polar triangles, each angle 
of one is measured by the supplement 
of the corresponding side of the other. 

Thus: 

A = 180° - a', B = 180° - &', C = 180° - c', 
a = 180° - A', b = 180° - B', c = 180° - C 

This result is of great importance ; for if any general 
equation be established between the sides and angles of a 
spherical triangle, it holds of course for the polar triangle 
also. Hence, by means of the above formulas any theorem of 
a spherical triangle may be at once transformed into another 
theorem by substituting for each side and angle respectively 
the supplements of its opposite angle and side. 

If a spherical triangle has one right angle, it is called a 
right triangle ; if it has two right angles, it is called a bi- 
rectangular triangle ; and if it has three right angles, it is 
called a tri-rectangular triangle. If it has one side equal to 
a quadrant, it is called a quadrantal triangle ; and if it has 
two sides equal to a quadrant, it is called a bi-quadrantal 
triangle. 




270 



SPHERICAL TRIGONOMETRY. 



Note. — It is shown in geometry that a spherical triangle may, in general, be 
constructed when any three of its six parts are given (not excepting the case in 
which the given parts are the three angles). In spherical trigonometry we investi- 
gate the methods by which the unknown parts of a spherical triangle may be com- 
puted from the above data. 

EXAMPLES. 

1. In the spherical triangle whose angles are A, B, C, 
prove 

B + C-A<tt (1) 

C+A-B<tt (2) 

A + B - C < 7T . . (3) 

2. If C is a right angle, prove 

A + B < f 7T (1), and A - B < ^ (2). 

It 

3. The angles of a triangle are A, 45°, and 120° ; find the 
maximum value of A. Ans. A < 105°. 

4. The angles of a triangle are A, 30°, and 150° ; find the 
maximum value of A. Ans. A < 60°. 

5. The angles of a triangle are A, 20°, and 110° ; find the 
maximum value of A. Ans. A < 90°. 

6. Any side of a triangle is greater than the difference 
between the other two. 



RIGHT SPHERICAL TRIANGLES. 

185. Formulae for Right Triangles. — Let ABC be a 

spherical triangle in which C is 
a right angle, and let be the 
centre of the sphere ; then will 
OA, OB, OG be radii: let a, b, c 
denote the sides of the triangle O* 
opposite the angles A, B, C, re- 
spectively ; then a, b, and c are 
the measures of the angles BOC, 
COA, and AOB. 







RIGHT SPHERICAL TRIANGLES. 271 

From any point D in OA draw DE _L to OC, and from E 
draw EF _L to OB, and join DF. Then DE is _L to EF (Geom. 
Art. 537). Hence (Geom. Art. 507), 

DF is 1_ to OB ; .-. Z DFE = Z B . . (Art. 183) 

AT OF OF OE , , • . ~ N 

Now = : that is, cos c = cos a cos b . . (1) 

OD OE OD' ' v ; 

DE = DEJDF h . , s in5 = sinBsinc . . (2) 
OD DF OD' v ' 

Interchanging a's and 6's, sin a = sin A sin c . . (3) 

— = — . — : that is, tan a= cos B tan c . . (4) 
OF DF OF' J v ' 

Interchanging a's and Us, tan b = cos A tan c . . (5) 

= ; that is, tan b = tan B sin a . . (6) 

OE EF OE' w 

Interchanging a's and &'s, tan a = tan A sin b . . (7) 
Multiply (6) and (7) together, and we get 

tanAtanB = — = -1- , by (1) 

cos a cos b cos c 

.♦. cos c = cot A cot B (8) 

Multiply crosswise (3) and (4), and we get 
sin a cos B tan c = tan a sin A sin c. 

.-. cosB = = sin A cos b, by (1) ... (9) 

cos a J v J v J 

Interchanging a's and Us, 

cos A = sin B cos a (10) 

Sch. By these ten formulae, every case of right triangles 
can be solved ; for every one of these ten formulae is a dis- 
tinct combination, involving three out of the five quantities, 
a, b, c, A, B, and there can be but ten combinations in all. 
Hence, any two of the five quantities being given and a 
third required, that third quantity may be determined by 
some one of the above ten formulae. 




272 SPHERICAL TRIGONOMETRY. 

186. Napier's Rules. — The ten preceding formulae, which 
may be found difficult to remember, have been included 
under two simple rules, called after their inventor, Napier's 
Rules of the Circular Parts. 

Let ABC be a right spherical triangle. Omit the right 
angle C. Then the two sides a 
and 6, which include the right 
angle, the complement of the 

hypotenuse c, and the comple- ^yf \b 

ments of the oblique angles A 

and B, are called the circular parts ' -<C — JC 

of the triangle. Thus, there are 

Jive circular parts, arranged in the figure in the following 

order : a, b, co. A, co. c, co. B. 

Any one of these five parts may be selected and called 
the middle part; then the two parts next to it are called 
adjacent paints, and the remaining two parts are called oppo- 
site parts. Thus, if co. A is selected as the middle part, 
then b and co. c are the adjacent parts, and a and co. B are 
the opposite parts. 

Then Napier's Bules are : 

(1) TJie sine of the middle part equals the product of the 
tangents of the adjacent parts. 

(2) The sine of the middle part equals the product of the 
cosines of the opposite parts. 

Note 1. — It will assist the student in remembering these rules to notice the 
occurrence of the vowel i in sine and middle, of the vowel a in tangent and adjacent, 
and of the vowel o in cosine and opposite. 

Napier's Rules* may be made evident by taking in detail each of the five parts as 
middle part, and comparing the equations thus found with the formulae of Art. 185. 

Thus, let co. c be the middle part. The rules give 

ein(co. c) = tan(co. A) tan(co. B) ; .*. cos c = cot A cot B (8) 

sin {co. c)= cos a cos b; .'. cos c = cos a cos b (1) 

co. B the middle part. 

ein(co. B) = tan a tan(co. c) ; .'. cos B = tan a cot c (4) 

Bin(co. B) = cos b cos(co. A) ; .*. cos B= cos b sin A (9) 

* While some find these rules to be useful aids to the memory, others question 
their utility. 



THE SPECIES OF THE PARTS, 273 

a the middle part. 

Bin a = tan b tan (co. B) ; .-. Bin a =tan b cot B (6) 

sin a = cos(co. A)cos(co. c) ; .\ sin a = sin A sin c (3) 

6 the middle part. 

sin 6 = tan a tan (co. A) ; .*. sin 6 = tan a cot A (7) 

sin o = cos(co. c)cos(co. B) ; .•. sin b =sin c sin B (2) 

co. A the middle part. 

sin(co. A)= tan b tan(co. c) ; .-. cos A= tan b cot c (5) 

8in(co. A)= cos a cos(co. B) ; .*. cos A= cos a sin B (10) 

Note 2. — In applying these rules it is not necessary to use the notation co.c, 
CO. A, co. B, since we may write at once cos c for sin (co. c), etc. 

187. The Species of the Parts. — If two parts of a spheri- 
cal triangle are either both less than 90° or both greater than 
90°, they are said to be of the same species. But if one part 
is less than 90° and the other part is greater than 90°, they 
are of different species. 

In order to determine whether the required parts are less 
or greater than 90°, it will be necessary carefully to observe 
their algebraic signs. If the required part is determined 
by means of its cosine, tangent, or cotangent, the alge- 
braic sign of the result will show whether it is less or 
greater than 90°. But when a required part is found in 
terms of its sine, it will be ambiguous, since the sines are 
positive in both the first and second quadrants. This 
ambiguity, however, may generally be removed by either 
of the following principles : 

(1) In a right spherical triangle, either of the sides con- 
taining the right triangle is of the same species as the opposite 
angle. 

(2) The three sides of a right spherical triangle (omitting 
bi-rectangular or tri-rectangular triangles) are either all 
acute, or else one is acute and the other two obtuse. 

The first follows from the equation 

cos A === cos a sin B, 



274 SPHERICAL TRIGONOMETRY. 




in which, since sinB is always positive (B < 180°), cos A 
and cos a must have the same sign; i.e., A and a must be 
either both < or both > 90°. 

The second follows from the equation 

cos c = cos a cos b. 

188. Ambiguous Solution. — When the given parts of a 
right triangle are a side and its opposite angle, the triangle 
cannot be determined. 

For two right spherical triangles ABC, A'BC, right 
angled at C, may always be 
found, having the angles A 
and A' equal, and BC, the 
side opposite these angles, 
the same in both triangles, 
but the remaining sides, AB, AC, and the remaining angle 
ABC of the one triangle are the supplements of the re- 
maining sides A'B, A f C, and the remaining angle A ? BC of 
the other triangle. It is therefore ambiguous whether 
ABC or A'BC be the triangle required. 

This ambiguity will also be found to exist, if it be 
attempted to determine the triangle by the equation 

sin b = tan a cot A, 

since it cannot be determined from this equation whether 
the side AC is to be taken or its supplement A'C. 

189. Quadrantal Triangles. — The polar triangle of a 
right triangle has one side a quadrant, and is therefore 
a quadrantal triangle (Art. 184). The formulae for quad- 
rantal triangles may be obtained by applying the ten 
formulae of Art. 185 to the polar triangle. They are as 
follows, c being the quadrantal side : 

cos C = — cos A cos B (1) 

sin B = sin b sin C (2) 



. 



EXAMPLES. 275 

sin A = sin a sin C (3) 

cos b = — tan A cot C (4) 

cos a = — tanB cot C (5) 

sin A = tan B cot b (6) 

sin B = tan A cot a (7) 

cos C = — cot a cot b (8) 

cos b = cos B sin a (9) 

cos a = cos A sin b (10) 

EXAMPLES. 

In the right triangle ABC in which the angle C is the 
right angle, prove the following relations : 

1. sin 2 a + sin 2 b — sin 2 c = sin 2 a sin 2 b. 

2. cos 2 A sin 2 c = sin 2 c — sin 2 a. 

3. sin 2 A cos 2 c = sin 2 A — sin 2 a. 

4. sin 2 A cos 2 b sin 2 c = sin 2 c — sin 2 b. 

5. 2 cos c = cos (a + b) + cos (a — b) . 

6. tan^(c + a) tan^(c — a) = tan 2 i&. 

. sin - = sin - cos - + cos - snr — 

2 2 2 2 2 

8. sin (c — 6) = tan 2 — sin (c + b). 

Li 

9. If 6 = c = -, prove cos a — cos A. 

10. If a = & = c, prove sec A = 1 + sec a. 

11. If c < 90°, show that a and 6 are of the same species. 

12. If c> 90°, a and b are of different species. 

13. A side and the hypotenuse are of the same or oppo- 
site species, according as the included angle <, or >-• 



276 



SPHERICAL TRIGONOMETRY. 




OBLIQUE SPHERICAL TRIANGLES. 

190. Law of Sines. — In any spherical triangle the sines 
of the sides are proportional to the sines of the opposite angles. 

Let ABC be a spherical triangle, the centre of the 
sphere ; and let a, b, c denote the 
sides of the triangle opposite the 
angles A, B, C, respectively. Then 
a, b, and c are the measures of the 
angles BOC, COA, and AOB. 

From any point D in OA draw 
DG _L to the plane BOC, and from 
G draw GE, GF J_ to OB, OC. 
Join DE, DF, and GO. Then DG 
is _L to GE, GF, and GO (Geom. Art. 487). Hence, DE is 
_L to OB, and DF JL to OC (Geom. Art. 507). 

.-. Z DEG = Z B, and Z DFG = Z C . . (Art. 183) 

In the right plane triangles DGE, DGF, ODE, ODF, 

DG = DE sin B = OD sin DOE sin B = OD sin c sin B, 

DG = DF sin C = OD sin DOF sin C = OD sin b sin C. 

.-. sin c sin B = sin b sin C ; 

or sin b : sin c : : sin B : sin C. 

Similarly, it may be shown that 

sin a : sin c : : sin A : sin C. 

sin b 



sin a 



sin c 



sin A sin B sin C 

Note. — The common value of these three ratios is called the modulus of the 
spherical triangle. 

Sch. In the figure, B, C, 6, c are each less than a right 
angle ; but it will be found on examination that the proof 
will hold when the figure is modified to meet any case 
which can occur. For example, if B alone is greater than 




LAW OF COSINES. 277 

90°, the point G will fall outside of OB instead of between 
OB and OC. Then DEO will be the supplement of B, and 
thus we shall still have sin DEG = sin B. 

191. Law of Cosines. — In any spherical triangle, the 
cosine of each side is equal to the product of the cosines of the 
other two sides, plus the product of the sines of those sides 
into the cosine of their included angle. 

Let ABC be a spherical triangle, the centre of the 
sphere, and a, b, c the sides of 
the triangle opposite the angles D 

A, B, C, respectively, Then ^^% \ V 

a = Z BOC, 
b = Z COA, 
c = Z AOB. 

From any point D in OA draw, in the planes AOB, 
AOC, respectively, the lines DE, DF J_ to OA. Then 

Z EDF = Z A (Art. 183) 

Join EF; then in the plane triangles EOE, EDF, we 
have 

EF 2 = OE 2 + OF 2 -20E.OFcosEOF . . (1) 

EF 2 = DE 2 + DF 2 -2DE.DFcosEDF . . (2) 

also in the right triangles EOD, FOD, we have 

OE^DE^OD 2 , and OF 2 -DF 2 = OD 2 . (3) 

Subtracting (2) from (1), and reducing by (3), and 
transposing, we get 

2 OE . OF cos EOF = 2 OD 2 + 2 DE . DF cos EDF. 

... cosEOF^^.^ + ^.^cosEDF, 
OF OE OF OE 

cos a = cos b cos c + sin b sin c cos A (4) 



278 SPHERICAL TRIGONOMETRY. 




By treating the other edges in order in the same way, 
or by advancing letters (see Note, Art. 96) we get 

cos b = cos c cos a + sin c sin a cos B . . (5) 
cos c = cos a cos b + sin a sin b cos C . . (6) 

Sch. Formula (4) has been proved only for the case in 
which the sides b and c are less than quadrants ; but it 
may be shown to be true when these sides are not less than 
quadrants, as follows : 

c •>■ 

(1) Suppose c is greater n — c— 

than 90°. Produce BA, BC B 
to meet in B', and put 
AB'=c', CB'=a', 

Then, from the triangle AB'C, we have by (4) 

cos a 1 = cos b cos c f + sin b sin c f cos B'AC, 

or cos(?r— a) = cos&cos(7r— c) + sin b sin(7r — c)cos(7r— A). 

.*. cos a = cos b cos c + sin b sin c cos A. 

(2) Suppose both b and c to 

be greater than 90°. Produce A / 
& A< 

AB, AC to meet in A', and put 

A'B = c', A'C == V. 

Then, from the triangle A'BC, we have by (4) 
cos a = cos b f cos c' + sin b' sin c f cos A f ; 
but b f = 7T — 6, c' = 7T — c, A' = A. 

.'. cos a = cos b cos c + sin 6 sin c cos A. 
The triangle AB'C is called the colunar triangle of ABC. 

192. Relation between a Side and the Three Angles. — 
In any spherical triangle ABC, 

cos A = — cos B cos C + sin B sin C cos a. 







RELATION BETWEEN SIDE AND ANGLES. 279 



Let A'B'C be the polar triangle of ABC, and denote its 
angles and sides by A', B', C, a', b', c' ; then we have by 
(4) of Art. 191 

cos a 1 — cos b' cos c r + sin b 1 sin c' cos A' ; 

but a' = i7 - A, V = 7T - B, c' = tt - C, etc. . (Art. 184) 

Hence, substituting, we get 

cos A== — cosB cos C +sinB sinC cos a .... (1) 

Similarly, 

cos B = — cos C cos A + sin C sin A cos b . . . . (2) 

cos C = — cos A cos B + sin A sin B cose .... (3) 

Rem. — This process is called " applying the formula to the polar triangle." By 
means of the polar triangle, any formula of a spherical triangle may be immediately 
transformed into another, in which angles take the place of sides, and sides of angles. 

193. To show that in a spherical triangle ABC, 

cot a sin b = cot A sin C + cos C cos b. 

Multiply (6) of Art. 191 by cos b, and substitute the 
result in (4) of Art. 191, and we get 

cos a = cos a cos 2 b + sin a sin b cos b cos C + si n 6 si 11 c cos A- 
Transpose cos a cos 2 b, and divide by sin a sin b ; thus, 

- . , , r, , sin c cos A 

cot a sin b = cos b cos C H ; 

sin a 

= cos b cos C + cot A sin C . (by Art. 190) 

By interchanging the letters, we obtain five other formulae 
like the preceding one. The six formulae are as follows : 

tcot a sin b = cot A sin C + cos C cos b . . . . (1) 



cot a sin c = cot A sin B + cos B cos c 
cot b sin a = cot B sin C + cos C cos a 
cot b sin c = cot B sin A + cos A cos c 
cot c sin a = cot C sin B + cos B cos a 
cot c sin b = cot C sin A + cos A cos b 



(2) 
(3) 
(4) 
(5) 
(6) 



280 



SPHERICAL TRIGONOMETRY. 



EXAMPLES. 

1. If a, b, c be the sides of a spherical triangle, a\ b\ c f 
the sides of its polar triangle, prove 

sin a : sin 6 : sin c = sin a f : sin V : sin c'. 

2. If the bisector AD of the angle A of a spherical 
triangle divide the side BC into the segments CD == b\ 
BD == c f , prove 

sin b : sin c = sin b ! : sin c f . 

3. If D be any point of the side BC ? prove that 

cot AB sin DAC + cot AC sin DAB = cot AD sin BAG. 
cot ABC sin DC + cot ACB sin BD = cot ADB sin BC. 

4. If a, fi, y be the perpendiculars of a triangle, prove that 

sin a sin a = sin b sin /3 = sin c sin y. 

5. In Ex. 4 prove that 



sin a cos a = Vcos 2 b + cos 2 c — 2 cos a cos b cos c. 

194. Useful Formulae. — Several other groups of useful 
formulae are easily obtained from those of Art. 191 ; the 
following are left as exercises for the student : 



sin a cos B = cos b sin c — sin b cos c cos A 
sin a cos C = sin b cos c — cos b sin c cos A 
sin b cos A = cos a sin c — sin a cos c cos B 
sin b cos C = sin a cos c — cos a sin c cos B 
sin c cos A = cos a sin b — sin a cos b cos C 
sin c cos B = sin a cos b — cos a sin 6 cos C 



(i) 

(2) 
(3) 
(4) 
(5) 
(6) 



FORMULA FOR THE HALF ANGLES. 281 

Applying these six formulae to the polar triangle, we 
obtain the following six : 



sin A cos b = cos B sin C + sin B cos C cos a 
sin A cos c = sin B cos C + cos B sin C cos a 
sinB cos a = cos A sin C + sin A cos C cos b 
sinB cos c = sin A cos C + cos A sin C cos b 
sin C cos a = cos A sin B + sin A cos B cos c 
sin C cos b = sin A cos B + cos A sin B cos c 



(7) 

(8) 

(9) 

(10) 

(11) 
(12) 



195. Formulae for the Half Angles. — To express the 
sine, cosine, and tangent of half an angle of a spherical 
triangle in terms of the sides. 

I. By (4) of Art. 191 we have 

cos A = cos a- cos b cose = ± _ 2 gin2 A ^ 4g 
sin b sin c 2 

2 A ^ cos a — cos b cos c 



.-. 2sin 2 ^ = l 

2 sin b sin c 

_ cos (b — c) — cos a ^ 
sin b sin c 

.-. sin 2A = sinfr(g+6--c)sinfr(a-6 + c) (Art. 45) 

2 sin b sin c 

Let 2s = a + b + c; so that s is half the sum of the sides 
of the triangle ; then 



a + b — c = 2(s — c), and a — b + c = 2(s — b). 



. 2 A sin (s — 6) sin (s — c) 

.\ snr — = * L — : — -• 

2 sin b sin c 



• sin A = / sin ( s ~ ^ sin ( S ~~ C "> (1) 

2 \ sin b sin c 



282 SPHERICAL TRIGONOMETRY. 

Advancing letters, 



. B /sin (s — c) sin (s — a) /ox 

sm-= A * — , J . — v L ... (2) 

2 \ sin c sin a 






cin C = J sin ( s ~ a) sin (& ' ~ 5 ) ... (3) 
2 \ sin a sin b 



II. 2 cos 2 - = 1 + cos A (Art. 49) 

_. i . cos a — cos 5 cose 

sin 6 sine 

i 

__ cos a — cos (6 + c) 
sin 6 sin c 

.\ cos 2 A = siQ -H a + & + <0 sin *-(& + c - a) 
2 sin b sin e 

_ sins sin(s — a) 
sin 6 sin c 



A /sins sin (s — a) //IX 

.-. COS— =\ ; ^ L (4) 

2 \ sin 6 sine 
Advancing letters, 

B /sins sin (a — b) /£ *\ 

COS — =a ; *S L (5) 

2 \ sine sm a 

C /sin s sin (s — c) /£} \ 

cos — =-v/ : ^ '- (6) 

2 \ sin asm 6 

III. By division, we obtain 

tan A = /sin^ ^&)sin( 8 -c) . . . (7) 
2 \ sins sin (s — a) 

ta B EMEiES ... (8) 
2 \ sins sin (s-6) 



tan^=J sin < s - a > sin ^- 6) ... (9) 

2 \ sins sin (s — c) 



FORMULAS FOR THE HALF ANGLES. 283 

Sch. The positive sign must be given to the radicals in 
each case in this article, because J A, ^B, i-C are each less 
than 90°. 

n -i , A, B sill (s — c) ,-,/xv 

Cor. 1. tan — tan— = ^ L (10) 

2 2 sins v y 

, B, C sin(s — a) /i . fX 

tan — tan— = ^ '- (11) 

2 2 sins v J 

tang tan A = sin( f -6) (12) 

2 2 sins v ' 

A A 

Cor. 2. Since sin A = 2 sin — cos — , 

2 2 ? 

. A 2Vsins sin(s — a) sin(s — &) sin(s — c) /1Q \ 

sin 6 sine 

= , 2n . (14) 

sin b sine 

where n 2 = sin s sin (s — a) sin (s — b) sin (s — c). 

EXAMPLES. 

-. -o .9* 1 — cos 2 a— cos 2 6 — cos 2 c-f-2cosacos&cosc 

1. Prove snrA= — — ^ 

sin-& sire 



1 



sin 2 6 sin 2 c 
where 4 n 2 = 1 — cos 2 a — cos 2 6 — cos 2 c + 2 cos a cos b cos c. 

2. Prove cose = cos(a+ &)sin 2 — |- cos (a — b) cos 2 — 

-r> • A . B . C sin(s— a) sin(s — &)sin(s— c) 

3. Prove sm — sm— sm — = * { * L * '-< 

2 2 2 sinasin6sinc 






m p cos A -f cos B _ sin (a + b) 

1 — cos C sin c 

k t> ^cosA + cosB . , , N . A 

5. Prove 2 -sm (a — 6) smc = 0. 

1 — cos C 

a -d cos A — cos B sin (a ~ b) 

6. Prove = * -• 

1 + cos C sin c 



284 SPHEBICAL TRIGONOMETRY. 

196. Formulae for the Half Sides. — To express the sine, 
cosine, arid tangent of half a side of a spherical triangle in 
terms of the angles. 

By (1) of Art. 192, we have 

COS A + COS B COS C i • 2& /A ac\\ 

cosa = — = 1 — 2sm 2 - . (Art. 49) 

sinBsinC 2 v ' 

• o s i n 2 a — cos A + cos (B + C) 
2 sin B sin C 

.-. zitf a = cosj(A+B+C)cosi(B+C-A) (Art> 45) 
2 sin B sin C 

Let 2S = A + B + C; then B + C - A = 2 (S - A). 

Proceeding in the same way as in Art. 195, we find the 
following expressions for the sides, in terms of the three 
angles : 

■ a_ I cos S cos (S — A) ,... 

Sm 2~ \ sin B sin C ^ ' 

.6 r^EmMEM ( 2 ) 

2 \ sinCsinA *• ' 

sin £ = J_^Scos(S-C) (g) 

2 \ sinAsinB v ' 

en- a = l / cos (S - B) cos (S - C) m 

2 \ sinBsinC K ' 

C08» = /cos(S-C)cos(S-A) 

2 \ sinCsinA V J 

cos£ = JcQ8(S-A)cos(S-B); (g) 

2 \ sinAsinB v ' 

tang = J cosScos(S-Ar2 . . . . ( 7 ) 

2 \ cos (S - B) cos (S - C) 



FORMULAE FOR THE HALF SIDES. 285 



tan* = J cosScos(S-B) ' 

2 V cos(S-C)cos(S-A) w 



tan£ = J cosScos(S-C)_ (0) 

2 \ cos(S-AHosfS-B) V ; 



/Sc/i. 1. These formulae may also be obtained immediately 
from those of Art. 195 by means of the polar triangle. 

Sch. 2. The positive sign must be given to the above 

radicals, because -, -, -, are each less than 90°. 

' 2 2 2 

Sch. 3. These values of the sines, cosines, and tangents 
of the half sides are always real. 

For S is > 90° and < 270° (Art. 184), so that cos S is 
always negative. 

Also, in the polar triangle, any side is less than the sum 
of the other two (Art. 184) . 

.\ 7T — A < 7T — B + 7T — C. 

.\ B + C-A<tt. 

.-. cos (S — A) is positive. 
Similarly, cos (S — B) and cos (S — C) are positive. 



Cor. Since sin a = 2 sin - cos -, 

2 2 

sin a== 2 ^'- cos Scos(S - A)cos(S - B)cos(S-C) (1Q v 
' Sma sin B sin C ^ ' 

2N 






sin B sin C 



where N= V— cos S cos (S — A) cos (S — B) cos (S — C). 



286 , SPHERICAL TRIGONOMETRY. 



EXAMPLES. 



1. Prove cosC=— cos(A+B)cos 2 - — cos(A— B) sin 2 - 

^ . a . b . c — N cos S 

2. Prove sm - sin - sin - = 









2 2 2 sinAsinBsinC 
where 1ST = V— cos S cos (S — A) cos (S — B) cos (S — C). 

197. Napier's Analogies. 

Let m = ?H^A = ^? (Art. 190) (1) 

sin a sin b 

sin A + sin B , K -, , \ /0 \ 

==— ^ , (Algebra) (2) 

sin a + sin 6 v & y v ' 

sin A — sin B /ON 

or = - — ■ ... (3) 

sm a ~ sm b 

cos A + cos B cos C = sin B sin C cos a (Art. 192) 

= m sin C sin b cos a, by (1 ) (4) 

and cos B + cos C cos A = sin C sin A cos b 

= m sin C sin a cos b . . (5) 

,\ (cos A + cosB)(l + cosC) = msinC sin (a + b), (6) 

from (4) and (5) 
Dividing (2) by (6), 

sin A + sin B _ sin a + sin b 1 +cosC 
cos A + cos B sin (a + 6) sinC 

.-. tan i ( A + B) = ^si(a-b) C (7) 

(Arts. 45, 46, and 49) 

Similarly, tan \ (A - B) = s | n * [ a ~ h) cot § . . (8) 






DELAMBBE'S ANALOGIES. 287 

Writing ir — A for a, etc., by Art. 184, we obtain from 
(7) and (8) 

. ,/ . 7X cos-HA — B) , c /rk x 

tan -J- (a + b) = ^ z tan-. ... (9) 

" V ; cos+(A + B) 2 V ; 

tan* (a - 6) = sin ^( A ~ B ) tan- .... (10) 
~ v ; sin£(A + B) 2 v ; 

ScJi. The formulae (7), (8), (9), (10) are known as 
Netpiei^s Analogies, after their discoverer. The last two 
may be proved without the polar triangle by starting with 
the formulae of Art. 191. 

Cor. In any spherical triangle ichose parts are positive, 
and less than 180°, the half-sum of any two sides and the half- 
sum of their opposite angles are of the same species. 

C 
For, since cos -i- (a — b) and cot— are necessarily positive, 

therefore by (7) tan -J- (A + B) and cos J (a + b) are both 
positive or both negative. 

.-. J (A + B) and J (a + b) are both > or both < or both 
= 90°". 

198. Delambre's (or Gauss's) Analogies. 

sin£(A + B) 

• A B , A . B 

= sin — cos — h cos — sin — 

2 2 2 2 



-v 



sin (s — b) sin (s — c) /sin s sin (s — b) 
sin b sin c V sin c sin a 






+ / sin s sin (s -a) / sin(s-c) sin (g - a) .^ ^ 
Af sin 6 sin c V sin c sin a 

_ sin (s — b) + sin (s — a) /sin s sin (s — c) 
sin c \ sin a sin 5 

cos - (Arts. 45 and 195) 



c 2 

cos- 

2 



288 SPHERICAL TRIGONOMETRY. 

c C 

.\ sin-|- (A + B) cos - = cos-|- (a — b) cos- . . . (1) 

2 2 

Similarly, we obtain the following three equations : 

c C 

sin -J- (A — B) sin-= sin -J (a — b) cos - ... (2) 

2 2 

c C 

cos % (A + B) cos - = cos \ (a + b) sin - . . . (3) 

2 2 

c C 

cos^(A — B) sin- = sin -J- (a + b) sin — ... (4) 

2 2 

Sch. 1. When the sides and angles are all less than 180°, 
both members of these equations are positive, except in (3). 

Sch. 2. Napier's analogies may be obtained from De- 
lambre's by division. 

Note. — Delambre's analogies were discovered by him in 1807, and published in 
the Connaissance des Temps for 1809, p. 443. They were subsequently discovered 
independently by Gauss, and published by him, and are sometimes improperly 
called Gauss's equations. Both systems may be proved geometrically. The 
geometric proof is the one originally given by Delambre. It was rediscovered by 
Professor Crofton in 1869, and published in the Proceedings of the London Mathe- 
matical Society, Vol. III. [Casey's Trigonometry, p. 41]. 



EXAMPLES. 

In the right triangle ABC, in which C is the right angle, 
prove the following relations in Exs. 1-45 : 

1. sin 2 a cos 2 5 = sin (c + b) sin (c — b). 

2. tan 2 a : tan 2 fr = sin 2 c — shi 2 b : sin 2 c — sin 2 a. 

3. cos 2 a cos 2 B = sin 2 A — sin 2 a. 

4. cos 2 A + cos 2 c = cos 2 A cos 2 c + cos 2 a. 

5. sin 2 A — cos 2 B = sin 2 a sin 2 B. 









6. If one of the sides of a right triangle be equal to the 
opposite angle, the remaining parts are each equal to 90° 









EXAMPLES. 289 

7. If the angle A of a right triangle be acute, show that 
the difference of the sides which contain it is less than 90°. 

-r» , B sin (s — <x) 

8. Prove tan — = * L - 

2 sin s 

9. Prove (1) 2n = sinasin6; (2) 2 1ST = sin a sin B. 

10. Prove sin 2 a sin 2 b = sin 2 a + sin 2 b — sin 2 c. 

11. Prove Uu^ = 8 ' m (°- b \ 

2 sin (c + b) 

12. Prove 2sin 2 - = sin 2 £(a + b) + silica- b). 

13. In a spherical triangle, if c = 90°, prove that 

tan a tan b + sec C = 0. 

14. In a spherical triangle, if c = 90°, prove that 

sin 2 p = cot cot cj>, 

where p is the perpendicular on c, and and <£ are the seg- 
ments of the vertical angle. 

15. Show that the ratio of the cosines of the segments 
of the base made by the perpendicular from the vertex is 
equal to the ratio of the cosines of the sides. 

16. If B be the bisector of the hypotenuse, show that 

sin 2 a + sin 2 b 



sin 2 B : 



4 cos 2 - 

2 



17. Prove tan S = cot - cot -. 

2 2 

18. Construct a triangle, being given the hypotenuse and 
(1) the sum of the base angles, and (2) the difference of 
the base angles. 

19. Given the hypotenuse and the sum or difference of 
the sides : construct the triangle. 



290 SPHERICAL TRIGONOMETRY. 



20. Given the sum of the sides a and b, and the sum 
of the base angles : solve the triangle. 



■ 



cm oi-l i.-u 4. A Vsin c + sin a + Vsin c — sin a 
21. Show that sin — == ! • 

2 2Vsinc 



22. sin 4- A; 



/sin_(c 
A/2 cos? 



J± 



b sin c 



23. cosiA=J 8in ( c +. & > - 
\ 2 cos 6 sm c 



/ 2 cos 6 sin c 

24. sin (a + 6) tan £ (A + B) = sin — 6) cot i (A — B). 
9 K • / a i t>\ __ cos a + cos b 





om \**--r^j 


1 + cos a cos 6 


26. 


sin (A B) - 


cos 6 — cos a 
1 — cos a cos 6 


27. 


cos (A + L) = 


sin a sin & 
1 + cos a cos 6 


28. 


cos (A B) - 


sin a sin 6 
1 — cos a cos 6 


29. 


sin 2 -=sin 2 ^ 

2 2 


o b , 9 a • o b 
cos 2 - + cos 2 -sm 2 -. 

2 2 2 


30. 


A 

sin (c — b) = sin (c + b) tan 2 — 


31. 


sin (a — 6) = 


sin a tan sin 6 tan — 

2 2 



32. sin (c — a) = cos a sin b tan — 

33. If ABC is a spherical triangle, right-angled at C, 
and cos A = cos 2 a, show that if A be not a right angle, 
b + c = \ it or § 7T, according as b and c are both < or 

both > £ 



EXAMPLES. 



291 






34. If «, ft be the arcs drawn from the right angle 
respectively perpendicular to and bisecting the hypotenuse 
c, show that 

sin 2 - (1 + sin 2 a) = sin 2 /3. 

35. In a triangle, if C be a right angle and D the middle 
point of AB, show that 

4 cos 2 - sin 2 CD = sin 2 a + sin 2 b. 

2 

In a right triangle, if p be the length of the arc drawn 
from the right angle C perpendicular to the hypotenuse 
AB, prove : 

36. cot 2 p = cot 2 a + cot 2 b. 

37. cos 2 p = cos 2 A + cos 2 B. 

38. tan 2 a = T tan a' tan c. 

39. tan 2 6 = ± tan 6' tanc. 

40. tan 2 a : tan 2 b = tan a r : tan b \ 

41. sin 2 j9 = sin a f sin b f . 

42. sinp sin c = sin a sin b. 

43. tan a tan & = tan c sin j?. 

44. tan 2 a + tan 2 6 = tan 2 c cos 2 p. 

45. cot A : cot B = sin a' : sin b'. 

In the oblique triangle ABC, prove the following : 

46. If the difference between any two angles of a tri- 
angle is 90°, the remaining angle is less than 90°. 

47. If a triangle is equilateral or isosceles, its polar tri- 
angle is equilateral or isosceles. 

48. If the sides of a triangle are each ^, find the sides 
of the polar triangle. 




292 SPHERICAL TRIGONOMETRY. 

49. If in a triangle the side a = 90°, show that 

cos A + cos B cos C = 0. 

50. If 6 and 6 f are the angles which the internal and ex- 
ternal bisectors of the vertical angle of a triangle make with 
the base, show that 

cos A ~ cos B n .. cos A + cos B 
cos 6 = ^ y and cos 0' = 

2 cos- 2sin- 

2 . 2 

cos .A 

51. Given the base c and ■ = — cos C : find the locus 

o , ., , cos B 

oi the vertex. 

52. Prove 4N 2 = 1 - cos 2 A - cos 2 B - cos 2 C 

— 2 cos A cos B cos C. 

53. If p, q, r be the perpendiculars from the vertices on 
the opposite sides, show that 

(1) sin a ship = sin b sin q = sin c sin r = 2n. 

(2) sin A sin p = sinB sin q= sinC sinr = 2N". 

54. Prove 8 n s = sin 2 a sin 2 b sin 2 c sin A sin B sin C. 

~~ -p. sin 2 A + sin 2 B + sin 2 C 1 + cos A cosB cosC 

55. Prove , \ . J , a -=-~- 7 

snra + snro + sure 1 — cos a cos b cose 

56. If Z be the length of the arc joining the middle point 

of the base to the vertex, find an expression for its length 

in terms of the sides. . , cos a + cos b 

Arts, cos 1 = 

2cos- 

2 

57. If CD, CD' are the internal and external bisectors of 
the angle C of a triangle, prove that 

L ~n cot a + cot 6 _ cota~cot& 
cot CD = — , and cot CD' = -^ 

2 cos — 2 sin — 

2 2 






EXAMPLES. 293 

58. Show that the angles and 6\ made by the bisectors 
of the angle C in Ex. 55 with the opposite side c, are thus 
given : 

cot a — cot b . „^ 
cot 6 = pj — sm CD, 

2sin~ 



,,. cot a + cot b . . 

cot f = ~ — sm CD'. 

2cos- 



59. Show that the arc intercepted on the base by the 
bisectors in Ex. 55 is thus given : 

j-tv™ sin 2 A — sin 2 B 

COt DD f = ; ; r-— •• 

2 sin A smB smC 

60. Prove that 

cos 2 6 — cos 2 c __ cos 2 c — cos 2 a 

cos b cot B — cos c cot C cos c cot C — cos a cot A 

cos 2 a — cos 2 b 



cos a cot A — cos b cos B 



61. If s and s f are the segments of the base made by the 
perpendicular from the vertex, and m and m f those made 
by the bisector of the vertical angle, show that 

s — s f , m — m' , 2 a — b 
tan tan = tan- - 

2 2 2 

62. Prove 

sin6 sine +cos6 cose cos A = sinB sinC — cosB cos C cos a. 

63. Show that the arc I joining the middle points of the 
two sides a and b of a triangle is thus given : 

1 + cos a 4- cos b + cos c 

cos I = ; . 

A a b 
4 cos - cos - 



294 SPHERICAL TRIGONOMETRY. 

64. If the side c of a triangle be 90°, and 8 the arc drawn 
at right angles to it from the opposite vertex, show that 

cot 2 8= cot 2 A + cot 2 B. 

65. Prove that the angle <£ between the perpendicular 
from the vertex on the base and the bisector of the vertical 
angle is thus given : 



tan d> = cos l(a-b) t w A _ B n 
r cos £ (a + b) 2V ' 

66. In an isosceles triangle, if each of the base angles be 
double the vertical angle, prove that 

a ( , a 

cos a cos - = cos c + - 

67. If a side c of a triangle be 90°, show that 

(1) cot a cot b + cos C = 0. 

(2) cos S cos (S - C) + cos (S - A) cos (S - B) = 0. 

68. In any triangle prove 

cos a — cos b sin (A — B) _ * 
1 — cos c sin C 

69. tan-J-(A + B):tan|(A-B) 

= tan ^ (<x + b) : tan |- (a — b). 

70. tan \ (A + a) : tan J ( A — a) 

= tani (B + b) : tan-J- (B - 6). 

71. If the bisector of the exterior angle, formed by pro- 
ducing BA through A, meet the base BC in D f , and if BD 
== c", CD' = b", prove 

sin b : sin c = sin b" : sin c". 

72. If D be any point in the side BC of a triangle, prove 

sinBD __ sin BAD # sin C 
sin CD "" sin CAD " sinB* 



EXAMPLES. 295 

73. If A = a, show that B and b are either equal or 
supplemental, as also C and c. 

74. If A = B + C, and D be the middle point of a, show 
that a = 2 AD. 

75. When does the polar triangle coincide with the 
primitive triangle ? 

76. If D be the middle point of c, show that 

cos a + cos 6 = 2 cos - cos CD. 

2 

77. In an equilateral triangle show that 

(1) 2 cos -sin — = 1. 
W 2 2 

(2) tan 2 - + 2cosA = l. 

78. If 6 + c = w, show that sin2B + sin2C = 0. 

79. Show that 

sin b sin c + cos b cos c cos A = sin B sin C — cos B cos C cos a. 

80. If D be any point in the side BC of a triangle, show 
that 

cos AD sin a = cos c sin DC + cos b sin BD. 

81. Prove cos 2 -=cos 2 -J-(a+6)sin 2 — + cos 2 |-(a — 6) cos 2 — 

82. " sin 2 ^= sin 2 ! (a +&; sin 2 ^ + sin 2 ^ (a -6) cos 2 -. 

83. u sin s sin (s — a) sin (s — b) sin (s — c) 
= £(1 — cos 2 a — cos 2 b — cos 2 c + 2 cos a cos b cos c). 

84. If AD be the bisector of the angle A, prove that 

A 

(1) cos B + cos C = 2 sin - sin ADB cos AD. 

A 

(2) cos C - cos B = 2 cos — cos ADB. 

v ' 2 



296 SPHERICAL TRIGONOMETRY. 

85. Prove cos a sin b = sin a cos b cos C + cos A sin c. 

86. " sin C cos a — cos A sin B + sin A cos B cos C. 

87. In a triangle if A = £, B = £ C = ~, show that 

D O 2t 

a + b + c = -• 

2 

00 -^ . /r , AN 1 + cos a — cos 6 — cose 

88. Prove sm (S — A) = ; 

A a . b > c 
4 cos - sm - sin - 

2 2 2 

89. If 8 be the length of the arc from the vertex of an 

isosceles triangle, dividing the base into segments a and ft, 

prove that 

, a , j8 , a + 8 , a — 8 
tan - tan £ = tan — ■ — tan • 

2 2 2 2 

90. If b — c, show that 

. a A 

sm - cos — 
2 2 

sin 6 = r , and sin B = 

. A a 

sm — cos - 

2 .2 

91. If AB, AC be produced to B', C, so that BB f , CC f 
shall be the semi-supplements of AB, AC respectively, 
prove that the arc B'C will subtend an angle at the centre 
of the sphere equal to the angle between the chords of 
AB, AC. 



PRELIMINARY OBSERVATIONS. 297 






CHAPTER XI. 
SOLUTION OF SPHEEIOAL TEIANGLES." 

199. Preliminary Observations. — In every spherical tri- 
angle there are six elements, the three sides and the three 
angles, besides the radins of the sphere, which is snpposed 
constant. The solution of spherical triangles is the process 
by which, when the valnes of any three elements are given, 
we calculate the values of the remaining three (Art. 184, 
Note). 

In making the calculations, attention must be paid to the 
algebraic signs of the functions. When angles greater than 
90° occur in calculation, we replace them by their supple- 
ments ; and if the functions of such angles be either cosine, 
tangent, cotangent, or secant, we take account of the change 
of sign. 

It is necessary to avoid the calculation of very small 
angles by their cosines, or of angles near 90° by their sines, 
for their tabular differences vary too slowly (Art. 81). It 
is better to determine such angles, for example, by means 
of their tangents. 

We shall begin with the right triangle ; here two ele- 
ments, in addition to the right angle, will be supposed 
known. 

SOLUTION OF RIGHT SPHERICAL TRIANGLES. 

200. The Solution of Right Spherical Triangles presents 
Six Cases, which may be solved by the formulae of Art. 
185. If the formula required for any case be not remem- 
bered, it is always easy to find it by Napier's Rules (Art. 



298 



SPHERICAL TRIGONOMETRY. 



186). In applying these rules, we must choose the middle 
part as follows : 

When the three parts considered are all adjacent, the 
one between is, of course, the middle part. When only- 
two are adjacent, the other one is the middle part. 

Let ABC be a spherical triangle, 
right-angled at C, and let a, 6, c 
denote the sides opposite the angles 
A, B, C, respectively. 

We shall assume that the parts 
are all positive and less than 180° 
(Art. 182). 




201. Case I. — Given the hypotenuse c and an angle A ; to 
find a, 6, B. 

By (3), (5), and (8) of Art. 185, or by Napier's Eules, 
we have 

sin a = sin c sin A, 

tan b = tan c cos A, 

cotB = cos ctanA. 

Since a is found by its sine, it would be ambiguous, but 
the ambiguity is removed because a and A are of the same 
species [Art. 187, (1)]. B and b are determined imme- 
diately without ambiguity. 

If a be very near 90°, we commence by calculating the 
values of b and B, and then determine a by either of the 
formulae 

tan a = sin b tan A, tan a = tan c cos B. 

Check. — As a final step, in order to guard against numer- 
ical errors, it is often expedient to check the logarithmic 
work, which may be done in every case without the neces- 
sity of new logarithms. To check the work, we make up 
a formula between the three required parts, and see whether 



CASE II. 



299 



it is satisfied by the results. In the present case, when the 
three parts a, b, B have been found, the check formula is 

sin a = tan b cot B . . . . [ (6) of Art. 185] 
Ex. 1. Given c = 81° 29' 32", A = 32° 18' 17" ; find a, b, B. 



Solution. 



log sine =9.9951945 
log sin A =9.7278843 

log sin a =9.7230788 

.-. a =31° 54' 25". 

log cose =9.1700960 
log tan A =9.8009157 

log cot B = 8.9710117 

.-. B = 84° 39' 21".33. 



log tan c =10.8250982 
log cos A = 9.9269687 

log tan b =10.7520669 

.-. b = 79° 51' 48".65. 

Check. 

log tan b =10.7520669 
log cot B = 8.9710117 

log sin a = 9.7230786 



Ex. 2. Given c = 110° 46' 20", A = 80° 10' 30" ; find a, 6, B. 
Ans. a = 67° 5' 52". 7, b = 155° 46' 42".7, B = 153° 58' 24".5. 



202. Case II. — Given the hypotenuse c and a side a ; to 
find b, A, B. 

By (1), (3), (4) of Art. 185, or by Napier's Kules, we 

have 

, cos c • A sin a -r> tan a 
cos b = , sin A = — — , cos B 



cos a 



sm c 



tan c 



The check formula involves 6, A, B ; therefore, from (9) 
of Art. 185 we have 

cos B = sin A cos b. 

In this case there is an apparent ambiguity in the value 
of A, but this is removed by considering that A and a are 
always of the same species (Art. 187). 



300 



SPHERICAL TRIGONOMETRY. 



Ex. 1. Given c = 140°, a = 20° ; find b, A, B. 
Solution. 



log cose = 9.8842540- 
colog cos a = 0.0270142 

log cos* = 9.9112682- 

.-. b = 144°36'28".4. 

log tan a = 9.5610659 
colog tan c= 0.0761865- 

log cos B = 9.6372524- 

.\ B = 115°42'23".8. 



log sin a =9.5340517 
colog sine = 0.1919325 

log sin A =9.7259842 

.-. A = 32°8'48".l. 

Check. 
log sin A =9.7259842 



log cos b 
log cos B 



: 9.9112682 
: 9.6372524 






Ex. 2. Given c = 72° 30', a = 45° 15' ; find b } A, B. 

Ans. 6 = 64° 42' 52", A = 48° V 44".5, B = 71° 27' 15". 

203. Case III. — Given a side a and the adjacent angle B ; 
to find A, b, c. 

By (10), (6), (4) of Art. 185, we have 

cos A = cos a sin B, tan b = sin a tan B, cot c = cot a cos B. 

Check formula, cos A = tan b cot c. 

In this case there is evidently no ambiguity. 
Ex. 1. Givena = 31°20'45", B = 55°30'30"; find A, b, a 
Solution. 



log cos a =9.9314797 
log sin B =9.9160371 

log cos A = 9.8475168 

.-. A = 45° 15' 30".6. 

log cot a =0.2153073 
log cos B = 9.7530361 

log cot c =9.9683434 

.-. c =47° 5' 11". 



log sin a =9.7161724 
log tan B =0.1630010 

log tan b =9.8791734 

.-. b =37° 7' 50". 

Check. 
log tan b = 9.8791734 
log cote =9.9683434 



log cos A 



: 9.8475168 



CASE IV. 301 

Ex. 2. Given a = 112° 0'0", B = 152° 23'1".3; find A, 6, c. 
Ans. A = 100°, b = 154° V 26".5, c = 70° 18' 10".2. 

204. Case IV. — Given a side a and the opposite angle A ; 
to find b, c, B. 

By (7), (3), (10) of Art. 185, we have 

, , , A sin a . -o cos A 

sin b = tana cot A, smc = , sin B = — 

sin A cos a 

Check formula, sin b = sin c sin B. 

In this case there is an ambiguity, as the parts are deter- 
mined by their sines, and two values for each are in general 
admissible. But for each value of b there will, in general, 
be only one value for c, since c and b are connected by the 
relation cos c = cos a cos b (Art. 185); and at the same time 
only one admissible value for B, since cos c = cot A cot B. 
Hence there will be, in general, only two triangles having 
the given parts, except when the side a is a quadrant and 
the angle A is also 90°, in which case the solution becomes 
indeterminate. 

It is also easily seen from a figure that the ambiguity 
must occur in general (Art. 188). 

When a= A, the formulae, and also the figure, show that 
b, c, and B are each 90°. 

Ex. 1. Given a = 46° 45', A = 59° 12' ; find b, c, B. 
Solution. 



log tan a = 0.0265461 



-o 



logcot A = 9.7753334 
log sin b =9.8018795 
.-. b =39°19'23".o, 
or 140° 40' 36". 5. 



log sin a =9.8623526 
log sin A =9.9339729 



log sine =9.9283797 

.-. c =57° 59' 29", 
or 122° 0'31". 



302 



SPHERICAL TRIGONOMETRY. 



log cos A = 9.7093063 

log cos a =9.8358066 

log sin B =9.8734997 
.-. B = 48°21'28", 
or 131° 38' 32". 



Check. 

log sine =9.9283797 
log sin B =9.8734997 

log sin b =9.8018794 



Ex. 2. Given a =112°, A = 100°; find b, c, B. 

Arts. & = 154° 7'26".5, c = 70°18'10".2, B = 152°23' 1".3, 

or 25°52'33".5, or 109°41'49".8, or 27° 36' 58 ".7. 

205. Case V. — Given the two sides a and b ; to find A, 

B, c. 

B y CO, ( 6 )> 0-) of Art - 185 > we have 

cot A = cot a sin b, cot B = cot b sin a, cos c = cos a cos b. 
Check formula, cos c = cot A cot B. 

In this case there is no ambiguity. 

Ex. 1. Given a =54° 16', b = 33° 12'; find A, B, c. 

Ans. A = 68°29'53" ? B=38°52'26" ? c = 60°44'46 r . 
Ex. 2. Given a = 56° 34', 6 = 27° 18'; find A, B, c. 

^4rw. A = 73° 9' 13", B = 31° 44' 9", c = 60° 41' 9". 

206. Case VI. — Given the two angles A and B ; to find a, 
6, and c. 



By (10), (9), (8) 
cos A 



cos a = 



., cos b = , cos c = cot A cot B. 



sin B sin A 

Check formula, cos c = cos a cos b. 
There is no ambiguity in this case. 

Ex. 1. Given A = 74° 15', B = 32° 10' ; find a, b, c. 

Ans. a = 59° 20' 44", b = 28° 24' 54", c = 63° 21' 24".5. 

Ex. 2. Given A = 91° 11', B = 111° 11' ; find a, 6, c. 

Ans. a = 91° 16' 8", b = 111° 11' 16", c = 89° 32' 29". 






EXAMPLES. 



303 



207. Quadrantal and Isosceles Triangles. — Since the 
polar triangle of a quadrantal triangle is a right triangle 
(Art. 184), we have only to solve the polar triangle by the 
formulae of Art. 185, and take the supplements of the parts 
thus found for the required parts of the given triangle ; or 
Ave can solve the quadrantal triangle immediately by the 
formulae of Art. 189. # 

A biquadrantal triangle is indeterminate unless either 
the base or the vertical angle be given. 

An isosceles triangle is easily solved by dividing it into 
two equal right triangles by drawing an arc from the vertex 
to the middle of the base. 

The solution of triangles in which a + b = 7r, or A + B = w, 
can be made to depend on the solution of right triangles. 
Thus (see the second figure of Art. 191) the triangle B'AC 
has the two equal sides, a 1 and b, given, or the two equal 
angles, A and B', given, according as a + 6 = 7rorA + B = 7r 
in the triangle ABC. 

examples: 

Solve the following right triangles : 



1. 






3. 



4. 



Given c=32°34', 
find a=22°15'43", 

Given c = 69°25'll", 
find «=o0° 0' 0", 

Given c=55° 9 '32", 
find b = 51° 53', 

Given c=127°12', 
find 6=39° 6' 25", 

Given a = 118° 54', 
find A =95° 55' 2", 



a=44°44'; 
6 = 24° 24' 19", 

A = 54° 54' 42"; 
b =56° 50' 49", 

o=22°15' 7"; 

A=27°28'37'.5, 

a=141°ll'; 
A =128° 5' 54", 

B=12°19'; 

& = 10°49'17", 



B=50°8'21". 
B = 63°25'4". 
B = 73° 27' 11". 1. 
B=52°21'49". 
c= 118° 20' 20". 



* Quadrantal triangles are generally avoided in practice, but when unavoidable, 
they are readily solved by either of these methods. 



304 
6. 

7. 

8. 

9. 
10. 
11. 
12. 



SPHERICAL TRIGONOMETRY. 
B = 137° 24' 21"; 



Given a=29° 46' 8", 
findA=54° 1'16", 

Given a= 77° 21' 50", 
find o=28°14'31".l, 
or o'=151°45'28".9, 

Given (( = 68°, 
find &=25°52'33".5, 
or 6'=154°7'26".5, 

Given a =144° 27' 3", 
find A =126° 40' 24", 

Given a =36° 27', 
find A=46°59'43".3, 

Given A= 63° 15' 12", 
find a= 49° 59' 56", 

Given A= 67° 54' 47", 



b =155° 27 '54", 

A=83°56'40"; 

c=78°53'20", 

c'=101°6'40", 

A=80°; 

c=70°18'10".2, 
c'=109°41'49".8, 

b =32° 8' 56"; 
B=47°13'43", 

b =43° 32' 31"; 
B=57°59'19".2, 

B = 135° 33' 39"; 

b =143° 5' 12", 
B = 99° 57' 35"; 

6=100° 45', 



c= 142° 9' 13". 

B = 28°49'57".4, 
B'=151°10'2".6. 

B =27°36'58".7, 
B'=152°23'l".3. 

c= 133° 32' 26". 

c=54°20'. 

c=120°55'34". 

c=94°5'. 



find a= 67° 33' 27", 

13. Solve the quadrantal triangle in which 

c = 90°, A = 42°l', B = 121°20'. 
Ans. C = 67°16'22", b = 112°10'20", a = 

14. Solve the quadrantal triangle in which 

a = 174°12'49".l, b = 94° 8' 20", c = 90°. 
Ans. A = 175°57'10", B = 135°42'55", C = 135°£ 

SOLUTION OF OBLIQUE SPHERICAL TRIANGLES. 

208. The Solution of Oblique Spherical Triangles presents 
Six Cases ; as follows : 

I. Given two sides and the included angle, a, b, C. 
II. Given two angles and the included side, A, B, c. 
III. Given two sides and an angle opposite one of them, 
a, b, A. 



CASE I. 305 

IV. Given two angles and a side opposite one of them, 
A, B, a. 

V. Given the three sides, a, b, c. 
VI. Given the three angles, A, B, C. 

These six cases are immediately resolved into three pairs 
of cases by the aid of the polar triangle (Art. 184). 

For when two sides and the included angle are given, 
and the remaining parts are required, the application of 
the data to the polar triangle transforms the problem into 
the supplemental problem : given two angles and the 
included side, to find the remaining parts. 

Similarly, cases III. and IV. are supplemental, also V. 
and VI. 

The parts are all positive and less than 180° (Art. 182). 
The attention of the student is called to Art. 199. 

209. Case I. — Given two sides, a, b, and the included 
angle C \ to find A, B, c. 

By Napier's Analogies, (7) and (8) of Art. 197, 

tanHA + B) = cosHa - 6) cotg. 
2V } cos^(a + 6) 2 

tan i (A - B) = sinH*-&) cot « 

2V } sin £ (a + 6) 2 

These determine |-(A + B) and -J- (A — B), and there- 
fore A and B ; then c can be found by Art. 190, or by one 
of Gauss's equations (Art. 198). Since c is found from its 
sine in Art. 190, it may be uncertain which of two values is 
to be given to it : if we determine c from one of Gauss's 
equations, it is free from ambiguity. We may therefore 
find c from (3) of Art. 198. Thus 

cos I (A + B) cos- = cos ^ (a + b) sin-- 

Z JL 



306 



SPHERICAL TRIGONOMETRY. 



Check, tan ^ (a + b) cos ^ (A + B) = cos |(A — B) tan -- 
There is no ambiguity in this case. 

Ex.1. Given a = 43° 18', b = 19° 24', C = 74°22'; find 
A, B, c. 

ySWwta'on. 
$ (a + &) = 31° 21', i(a - b) = 11° 57', | C = 37° 11'. 



log cos i(a-b)= 9.9904848 
log sec i (a + b) = 0.0685395 

log cot ^=10.1199969 

8 2 



logtan|(A+B) =10.1790212 

.-. £(A + B) = 56°29'17" 
£(A-B) = 27°41' 0».5 

.-. A = 84° 10' 17".5 

B = 28° 48' 16".5. 
c = 41°35'48".5. 



log sin \ (a - b) = 9.3160921 
log cosec | (a + b) = 0.2837757 

log cot -=10.1199969 

8 2 

logtan|(A-B) = 9.7198647 

.-. £(A-B)=27°41'0".5. 

log cos i (a + b) = 9.9314605 
log sec 1(A + B) = 0.2579737 

log sin £= 9.7813010 

8 2 

log cos - = 9.9707352 

8 2 

.-. £ = 20°47'54".25. 
2 



Otherwise thus: Let fall the 
perpendicular BD, dividing the 
triangle ABC into two right tri- 
angles, BDA, BDC. Denote AD 
by m, the angle ABD by <f>, and A- 
BD by p. Then by Napier's Rules, 
we have 

cos C = tan (6 — m) cot a ; 

sin (& — m) = cot C t&np ; sin m = cot A tan p. 

.*. tan (6 — m) = tanacos C (1) 

and tan A sin m = tan C sin (6 — m) (2) 




case II. 397 

From (1) m is determined, and from (2) A is determined. 
In a similar manner B may be found. 

Also, from the same triangles, we have by Napier's Rules 

cos a = cos (b — m) cosp ; cos c = cos m cosp. 

.*. cos c cos (b — m) = cos m cos a, 

from which c is found. 

Note. — This method has the advantage that, in using it, nothing need be remem- 
bered except Xapier's Rules. 

If only the side c is wanted, it may be found from (4) of Art. 191, without pre- 
viously determining A and B. This formula may be adapted to logarithms by the 
use of a subsidiary angle (Art. 90). 

Ex.2. Given 6 = 120° 30' 30", c=70°20'20", A = 50°10'10"; 
find B, C, a. 

Arts. B = 135° 5' 28".8, C = 50° 30' 8".4, a = 69° 34' 56'. 

210. Case II. — Given two angles, A, B, and the included 
side c ; to find a, b, C. 

By Napier's Analogies (9) and (10) of Art. 197, 

tan | (a + b) = C0S K A - B ) tan £, 
2V ; cosi(A + B) 2 

tan|(q-^&)^ sin ^ A - -^ltan^ 
2V J sin£(A + B) 2 

from which a and b are found. 

The remaining part C may be found by (2) of Art. 198. 

C c 

sin | (a — b) cos — = sin ^ (A — B) sin — 

C 
Check, cos \ (a — b) cot — = cos | (a + b) tan \ ( A + B) . 

Li 

There is no ambiguity in this case. 



308 



SPHERICAL TRIGONOMETRY. 



Ex. 1. Given A=68°40', B = 56°20', c=84°30'; find a, b, C. 

Solution. 
£ (A + B) = 62° 30', £ (A - B) = 6° 10', § = 42° 15'. 



logcos£(A-B) = 9.9974797 
logsecJKA+B) = 0.3355944 

log tan - = 9.9582465 

& 2 . 



log tan £ (a + b) = 10.29l.3206 

.-. £(a + 5) = 62° 55' 9" 

fr(q-6) = 6° 16' 3 9" 

a = 69° 11' 48" 

b = 56° 38' 30". 
C = 97°19' 3".5. 



logsini(A-B) =9.0310890 
logcoseci(A+B) =0.0520711 

log tan -=9.9582465 

2 . 



log tan |(a - b) =9.0414066 

... i(a-6)=6°16'39". 

log sin i(A-B) =9.0310890 
log cosec i(a - 6) =0.9612050 

log sin £=9.8276063 

& 2 

log cos -=9.8199003 

& 2 

. H = 48° 39' 31|". 



Otherwise thus : Let fall the perpendicular BD (see last 
figure). Denote, as before, AD by m, the angle ABD by ty, 
and BD by p. Then by Napier's Bules, we have 

cos c = cot <£ cot A ; 

cos <j> = cot c tan p ; cos (B — <£) = cot a tan p. 

.: cot <f> = tan A cos c (1) 

tan a cos (B — <£) = cos <j> tan c (2) 

From (1) <t> is determined, and from (2) a is found. 
Similarly b may be found. 

Also, from the same triangles, we have 

cos C sin <£ = cos A sin(B — <f>), 

from which C is found. 



I 



CASE III. 309 

Ex. 2. Given 

A = 135° 5' 2S".G, C = 50° 30' 8".6, b = 69° 34' 56".2 ; 
find a, c, B. 

An*, a = 120° 30' 30", c = 70° 20' 20", B = 50° 10' 10". 

211. Case III. — Given two sides, a, b, and the angle A 
opposite one of them ; to find B, C, c. 

The angle B is found from the formula, 

sin B= 5^ sin A (Art. 190) (1) 

sin a 

Then C and c are found from Napier's Analogies, 



tan 



sin£(q-&) co t i(A-B) .... (2) 



c 

2 " sinJ(a + 6) 



Check, 



sin A _ sin B _ sin C 

sin a sin 6 sin c 



Since B is found from its sine in (1), it will have two 
values, if sin A sin b < sin a, and the triangle, in general, 
will admit of two solutions. 

When sin A sin b > sin a, there will be no solution, for 
then sin B > 1. 

In order that either of these values found for B may be 
admissible, it is necessary and sufficient that, when sub- 
stituted in (2) and (3), they give positive values for 

C c 

tan— and tan-, or which is the same tiling, that A — B and 
2 2 

a — b have the same sign. Hence we have the following 

Rule. — If both values of B obtained from (1) be such as 
that A — B and a — b have like signs, there are two complete 
solutions. If only one of the values of B satisfies this condi- 
tion, there is only one triangle that satisfies the problem, since 



310 



SPHERICAL TRIGONOMETRY, 



in this case C, or c > 180°. If neither of the values of B 
makes A — B and a — b of the same signs, the problem is 
impossible. 

This case is known as the ambiguous case, and is like the 
analogous ambiguity in Plane Trigonometry (Art. 116), 
though it is somewhat more complex. For a complete 
discussion of the Ambiguous Case, the student is referred 
to Todhunter's Spherical Trigonometry, pp. 53-58 ; McCol- 
lend and Preston's Spherical Trigonometry, pp. 137-143; 
Serret's Trigonometry, pp. 191-195, etc. 

Ex. 1. Given a=42°45', 6 = 47°15', A=56°30'; findB, C, a 






Solution. 



log sin b = 9.8658868 

colog sin a = 0.1682577 

log sin A = 9.9211066 

log sin B =9.9552511 

.\ B = 64°26'4", 

B' = 115° 33' 56". 



*(« + &) = 


45° 0' 0". 


l(a-6) = - 


- 2° 15' 0". 


i(A + B) = 


60° 28' 2". 


i(A-B)=- 


- 3° 58' 2". 


J(A + B') = 


86° 1'58". 


i(A-B') = - 


- 29° 31' 58". 



Since both values of B are such that A — B, A — B', and 
a — b, are all negative, there are two solutions, by the above 
Rule. 



(1) When B = 64° 26' 4". 

log sin £ (a - 6) =8. 5939483 - 
colog sin \ (a+ b) = 0.1505150 
log cot i(A - B) =1.1589413- 

log tan- =9.9034046 



C 



= 38° 40' 48". 



C = 77° 21' 36". 



log sin i(A + B) =9.9395560 
cologsin$(A-B)=1.1599832- 

log tan | (a - b) =8.5942832- 

log tan- =9.6938224 

° 2 



c 

2 : 



:26°17'40". 



c =53° 35' 20". 



CASE III. 



311 



(2) WhenB' = 115°33'56". 



log sin £(«-&) =8.5939483- 
colog sin | (a + 6 ) = 0.1505150 
log cot J(A-B') = 0.2467784- 

log tan --=8.9912417 

5 2 



C 



5°35'50i". 



•. c'=n°ir40£". 



logsin|(A + B') =9.9989581 
cologsin|(A-B') = 0.3072223- 

log tan|(a- 6) = 8.5942832- 

log tan -=8.9004636 

6 2 

.-. - = 4°32'47i". 

2 4 

.-. c'=9° 5'34|". 




^4ns. B = 64° 26' 4", C =77° 21' 36", c =53° 35' 20"; 
B' = 115° 35' 56", C' = H°11'40J-", c'= 9° 5'34£". 

Otherwise thus: Let fall the 
perpendicular CD; denote AD 
by m, the angle ACD by <f>, and 
CD by p. Then we have 

cos A = tan m cot b ; 

.% tan m = cos A tan b (1) 

cos b = cot A cot <£. .-. cot 4> = cos b tan A 

Again, 

cos a = cos (c — m) cosp ; cos 6 = cos m cos p. 

.-. cos (c — m) = cos a cos m -r- & . . 

Also, cos (C — cj>) = cot a tanp ; cos <£ = cot b tanp. 

•% cos (C — <f>) = cot a tan 6 cos <£ . . 

sin b 



(2) 



(3) 



Lastly, 



sinB : 



sin A 



sma 



(4) 

(5) 



The required parts are given by (1), (2), (3), (4), (5). 

Ex. 2. Given a = 73°49'38", 6=120° 53' 35", A = 88°52'42": 
find B, C, a 

Ans. B = 116° 44' 48", C = 116° 44' 48", c = 120° 55' 35". 



312 



SPHERICAL TRIGONOMETRY. 






212. Case IV. — Given two angles, A, B, and the side a 
opposite one of them ; to find b, c, C. 

This case reduces, by aid of the polar triangle, to the 
preceding case, and gives rise to the same ambiguities. 
Hence the same remarks made in Art. 211 apply in this 
case also, and the direct solution may be obtained in the 
same way as in Case III. Thus, 

The side b is found from the formula 
sinB 



sin b- 



sin A 



sin a (1) 



Then c and C are found from Napier's Analogies. 
c = cosHA + B) 
2 cos^-(A-B) 2 v ' 



C = cosHa-6) A B 

* cos! (a + 6) 



Check, 



2 

sin A 
sin a 



(2) 
(3) 



sin B _ sin C 
sin b 



sine 



Ex. 1. Given A=66°7'20", B=52°50'20", a=59°28'27"; 
find b, c, C. 



By (1) we find b ■ 



Solution. 
: 48° 39' 16", or 131° 20' 44". 



i(A + B)=59°28'50". 
i(A-B)= 6° 38' 30". 

logcosi(A + B)= 9.7057190 

cologcos|(A-B)= 0.0029244 

log tan | (a + b) = 10.1397643 



log tan -= 9.8484077 

8 2 

- = 35°11'50|". 



£(« + &) = 54° 3'51£". 
£ («_«.)= 5° 24' 35 i". 

log cos \ (a - b) =9.9980612 
colog cos f(a+ 6) =0.2314530 
log cot£ ( A + B ) = 9. 7704854 

log tan §=9.9999996 



C 

2' 



:45°. 



The second value of b is inadmissible (see Rule of Art. 
211), and therefore there is only one solution. 

Ana. b = 48° 39' 16", c = 70° 23' 41£", C = 90°. 






CASE V. 313 

Ex. 2. Given A = 110° 10', B = 133° 18', a = 147° 5' 32" ; 
find b, c, C. 

Ans. b = 155° 5' 18", c = 33° V 45", C = 70° 20' 50". 

213. Case V. — Given the three sides, a, 6, c ; to find the 
angles. 

The angles may be computed by any of the formulae of 
Art. 195 ; but since an angle near 90° cannot be accurately 
determined by its sine, nor one near 0° by its cosine (Art. 
151), neither of the first six formulae can be used with advan- 
tage in all cases. The formulae for the tangents however are 
accurate in all parts of the quadrant, and are therefore to 
be preferred for the solution of a triangle in which all three 
sides or all three angles are given. 

By (7) of Art. 195 we have 

tan - / sin(s-&)sin(s--c) 
2 \ sin s sin (s — a) 

— 1 / sin (s — a) sin (s — b) sin (s — c) 

sin(s — a) \ sins 

Since the part under the radical is a symmetric function 
of the sides, it is in the forinulse for determining all three 
angles A, B, C, and when once calculated, it may be utilized 
in the calculation of each angle. For convenience in com- 
putation, denote this term by tan r. Then 

tan r = v / sin ( * ~ a ^ sin ^ ~ h) sin (g ~ c ^ ; 
\ sins 

and (7), (8), (9) of Art. 195 become 

. A tanr ,-ts 

tan — = (1 ) 

2 sm(s-a) v y 

. B tanr /0 \ 

tan — = (2) 

2 sin(s-6) v J 

tang= . t / anr x (3) 

2 sin(s-c) v J 



314 



SPHERICAL TRIGONOMETRY. 



Check, 



sin A sin B sin C 



sin a 



sin b 



sine 



Ex.1. Given a = 46° 24', 6 = 67° 14*, c = 81°12'; find 

A, B, C. 

Solution. 



a= 46° 24' 
6= 67° 14' 
c = 81° 12' 



2s = 194°50' 

s= 97° 25', 

s-a= 51° V, 

s-b= 30° 11', 

s-c= 16° 13'. 



log sin (s-a) = 9.8906049 

log sin (s - 6) = 9.7013681 

log sin (s - c) = 9.4460251 

colog sin s = 0.0036487 

log tan 2 r = 9.0416468 

log tan r = 9.5208234. 



tan r =9.5208234 
sin (s-a)= 9.8906049 

tan -=9.6302185 

2 



A 

2 
A 



=23° 6*45". 
=46° 13' 30". 



tan r =9.5208234 
sin(s-6)=9.7013681 
B 



tan : 



B 



: 9.8194553 



=33° 25' 10". 



B =66° 50' 20". 



tanr= 9.5208234 
sin(s-c)= 9.4460251 

tan £ =10.0747983 



c 



=49° 54' 35". 



C =99° 49' 10". 



Ex. 2. Given a = 100°, b = 37° 18', c = 62° 46 f ; find A, 
B, C. 

Arts. A=176°15'46".56, B=2°17'55".08, C = 3°22'25".46. 

214. Case VI. — Given the three angles, A, B, C ; to find 
the sides. 

As in Art. 213, the formulae for the tangents are to be 
preferred. 

Putting tan R = \ - 

\c 



cos S 



/cos (S - A) cos (S - B) cos (S - C) 
we have, from (7), (8), (9) of Art. 196, 



tan - = tan R cos (S — A), 



CASE VI. 



315 



tan - = tan R cos (S — B), 



tan - = tan R cos (S — C), 
by which the three sides may be found. 

Check, smA = sinB = sinC. 

sin a sin b sin c 

Ex.1. Given A = 68° 30', B = 74° 20', C = 83°10'; find 
a, b, c. 

Solution. 



A= 68° 30' 
B = 74° 20' 

C = 83° 10' 
2S = 226° 0' 

log (- cos S) = 9.5918780 

log cos (S - A) = 9.8532421 

log cos (S - B) = 9.8925365 

log cos (S - C) = 9.9382576 

log tan 2 B = 9.9078418 

log tan E = 9.9539209* 



S = 113° 0' 
S_A= 44° 30' 
S-B= 38° 40' 
S - C = 29° 50' 



log tan ' 



■■ 9.8071630 



log tan -=9.8464574 

8 2 

log tan - = 9.8921785 

8 2 



a = 65° 21' 22£" 
6 = 70° 9' 9|", 
c = 75°55' 9". 



Check, 



sin a 



sin 6 



sine 



sin A sin B sin C 



Ex.2. Given A=59°55'10", B = 85°36'50°, C = 59°55'10"; 
find a, b, c. 

Ans. a = 129° 11' 40", b = 63° 15' 12", c = 129° 11' 40". 

* The necessary additions may be conveniently performed by writing log tan R on 
a slip of paper, and holding it successively over log cos (S — A), log cos (S — B), and 
logcos(S-C). 



316 



SPHERICAL TRIGONOMETRY. 



EXAMPLES. 
Solve the following right triangles : 



2. 

3. 

4. 

5. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 



Given c= 84° 20', 
find a= 35° 13' 4", 

Given c= 67° 54', 
find a= 39° 35' 51", 

Given c= 22° 18' 30", 
find a= 16°17'41", 

Given c=145°; 
find a= 13° 12' 12", 

Given c= 98° 6' 43", 
find a=137° 6', 

Given c= 46° 40' 12", 
find a= 26°27'23".8, 

Given c= 76° 42', 
find b= 70° 10' 13", 

Given c= 91° 18', 
find b= 94°18'53".8, 

Given c= 86° 51', 
find a= 86° 41' 14", 

Given c= 23° 49' 51", 
find 6= 19° 17', 

Given c= 97° 13' 4", 
find 6= 79°13'38".2, 

Given c= 37° 40' 20", 
find 6= 0°26'37".2, 



A= 35° 25'; 
b= 83° 3' 29", 

A= 43° 28'; 
6= 60°46'25|", 

A= 47° 39' 36"; 
6= 15° 26' 53", 

A= 23° 28'; 
6= 147° 17' 15", 

A=138°27'18"; 
b= 77° 51', 

A= 37° 46' 9"; 
b= 39°57'41".4, 

a= 47° 18'; 
A= 49° 2'24".5, 

a= 72° 27'; 
A= 72° 29' 48", 

b= 18° 1'50"; 
A= 88° 58' 25", 

a= 14° 16' 35"; 
A= 37°36'49".3, 

a=132°14'12"; 
A=131°43'50", 

a= 37° 40' 12"; 
A= 89°25'37", 



B = 85° 59' 1". 
B = 70°22'21". 
B= 44°33'53".4 
B= 109° 34' 33". 
B= 80° 55' 27". 
B= 62° 0' 4". 
B= 75°9'24".75J 
B= 94° 6'53".3J 
B= 18° 3' 32". 
B= 54°49'23".3. 
B= 81°58'53".3. 
B= 0°43'33", 



EXAMPLES. 



317 



Given a-- 
find A= 

Given a- 

findA: 

Given a- 
find A = 

Given a- 
find A = 

Given b- 
find B = 

Given b- 
find B = 

Given a= 
find 6= 
or b'= 

Given a- 
find b= 
or 6' = 

Given a= 
find 6= 
or b'= 

Given a- 
find b-- 
or 6'= 

Given 6= 
find a- 
or a'= 



. 82° 6', 
= 84° 34' 28", 

= 42° 30' 30", 
: 53° 50' 12", 

: 20° 20' 20", 
= 54°35'16".7, 

= 92°47'32"; 
= 92° 8' 23", 

= 54° 30', 
= 70° 17' 35", 

:155°46'42".7, 
= 153°58'24".5, 

: 35° 44', 
: 69° 50' 24", 
:110° 9' 36", 

: 129° 33', 
: 18° 54' 38", 
=161° 5' 22", 

= 21° 39', 

= 25°59'27".8, 
= 154° 0'32".2, 

-. 42° 18' 45", 
= 60° 36' 10", 
: 119° 23' 50", 

= 160°, 

= 39° 4' 50". 7, 

= 140° 55' 9".3, 



B= 43° 28'; 
b= 43° 11' 38", 

B= 53°10'30"; 
b= 42° 3'47", 

B= 38° 10' 10"; 
b= 15° 16' 50 ".4, 

B= 50° 2' 1"; 

b= 50°, 

A= 35° 30'; 
a= 30° 8'39".2, 

A= 80° 10' 30"; 
a= 67° 6'52".6, 

A = 37° 28'; 
c= 73° 45' 15", 
c'=106°14'45", 

A = 104° 59'; 
c=127° 2'27", 
c'= 52°57'33", 

A= 42°10'10"; 
c= 33°20'13".4, 
c'=146°39'46".6, 

A= 46°15'25"; 
c= 68°42'59", 

C'=:lll°17' 1", 

B = 150°; 
c=136°50'23".3, 
c'= 43° 9'36".7, 



84° 14' 57". 



c= 56° 49' 8". 



c= 25°14'38".2. 



c= 91°47'40". 



59°51'20". 



c= 110° 46' 20". 



B= 

B = 

B = 
B'= 

B = 

B'= 

B = 

B' = 

A= 

A'= 



= 77° 54', 
=102° 6'. 

: 23° 57' 19", 
=156° 2' 41". 

= 52°23' 2".8, 
:127°36'57".2. 

-. 69°13'47", 
=110°46'13". 

= 67° 9'42".7, 
=112°50'17".3. 



318 SPHERICAL 

24. Given a = 25° 18' 45", 
Ans. Impossible ; why? 

Given a= 32° 9' 17", 
find A= 49° 20' 17", 

Given a= 55° 18', 
find A= 66° 15' 6", 

Given a= 56° 20', 
find A= 56° 51' 7", 

Given a= 86° 40', 
find A= 88°11'57".8, 

Given a= 37° 48' 12", 
find A= 41° 55' 45", 

Given a =116°, 
find A= 97°39'24".4, 

Given A = 52° 26', 
find a= 36°24'34".5, 

Given A = 64° 15', 
find a= 54° 28' 53", 

Given A = 54° 1'15", 
find a= 29° 46' 8", 

GivenA= 46°59'42", 
find a= 36° 27', 

Given A = 55° 32' 45", 
find a= 54°41'35", 

Given A= 60°27'24".3, 
find a= 54°32'32".l, 



TRIGONOMETRY. 
A= 15° 58' 15". 

6= 32° 41'; 
B= 50° 19' 16", 

b= 39° 27'; 
B = 45° 1'31", 

b= 78° 40'; 
B= 80° 31' 48", 

b= 32° 40'; 
B = 32°42'37".8, 

b= 59°44'16"; 
B= 70° 19' 15", 

b= 16°; 
B= 17°41'39".9, 

B = 49° 15'; 
6= 34°33'40", 

B= 48° 24'; 
6= 42°30'47", 



B = 137° 24' 21"; 

6= 155° 27' 55", 

B= 57° 59' 17"; 
b= 43°32'37", 



B=101°47'56"; 
6=104°21'28", 

B= 57°16'20".2: 
b= 51°43'36".l, |c= 



c= 44°33'17". 

c= 63°55'21". 

c= 83°44'44£". 

c= 87°11'39".8. 

c= 66° 32' 6". 
c=114°55'20".4. 

c= 48° 29' 20". 

c= 64°38'38". 



c=142° 9' 12". 



54° 20' 3". 



EXAMPLES. 


319 


Solve the following quadrantal triangles : 




Given B = 74° 45', 


a= 


18° 12', 


c= 90°; 


find b= 85°17'15".5, 


A= 


17° 34' 2", 


C = 104° 31' 13". 


Given A =110° 47' 50", 


B= 


135°35'34''.5, 


c= 90°; 


find a=104°53' 0".8, 


b= 


133° 39' 47". 7, 


C = 104°41'37".2 


Solve the following oblique triangles : 




Given o= 73° 58', 


b= 


38° 45', 


C= 46° 33' 39"; 


find A =116° 8' 28", 


B = 


35° 46' 39", 


c= 51° I'll". 


Given a= 96° 24' 30", 


b= 


68° 27' 26", 


C= 84° 46' 40"; 


find A= 97° 53' 01", 


B = 


67° 59' 391", 


c= 87°31'37". 


Given a= 76°24'40", 


b = 


58° 18' 36", 


C = 116° 30' 28"; 


find A= 63° 48' 351", 


B = 


51° 46' 121", 


c=104°13'27". 


Given a= 86° 18' 40", 


6 = 


45° 36' 20", 


C = 120° 46' 30"; 


find A= 64°48'53f", 


B = 


40°23'15f", 


c= 108° 39' Hi". 


Given a= 88° 24', 


b= 


56° 48', 


C = 128°16'; 


find A= 65° 13' 3|", 


B= 


49° 27' 51", 


c= 120° 10' 52". 


Given a= 68° 20' 25", 


b= 


52° 18' 15", 


C=117°12'20"; 


find A= 56° 16' 15", 


B = 


45° 4' 41", 


c= 96°20'44". 


Given a= 88°12'20", 


b= 


124° 7' 17", 


C= 50° 2' 1"; 


find A= 63° 15' 12", 


B = 


132° 17' 59", 


c= 59° 4'25". 


Given a= 32°23'57", 


b= 


32° 23' 57", 


C= 66°49'17"; 


find A= 60° 53' 2", 


B = 


60° 53' 2", 


c= 34° 19' 13". 


Given 6= 99° 40' 48", 


c= 


100°49'30", 


A= 65° 33' 10"; 


find B = 95° 38' 4", 


C = 


97°26'29".l, 


a= 64°23'15".l 


Given A = 31° 34' 26", 


B= 


30° 28' 12", 


c= 70° 2' 3"; 


find a= 40° 1' 5\", 


b= 


38° 31' 3£", 


C=130° 3' 50". 



320 



SPHERICAL TRIGONOMETRY. 



49. 


GivenA=130°5'22".4, 


B = 32°26'6".41, 


c= 51° 6'11".6 




find a- 84°14'29", 


b= 44° 13' 45", 


C= 36° 45' 26". 


50. 


GivenA= 96°46'30", 


B= 84° 30' 20", 


c=126°46'; 




find a=102°21'42", 


b= 78° 17' 2", 


C= 125° 28' 131". 


51. 


Given A = 84° 30' 20", 


B= 76° 20' 40", 


c=130°46'; 




find a- 94° 34' 521", 


b= 76° 40' 481", 


C= 130° 51' 331". 


52. 


Given A= 107° 47' 7", 


B= 38°58'27", 


c= 51° 41' 14"; 




find a= 70° 20' 50", 


b= 38° 27' 59", 


C= 52° 29' 45". 


53. 


GivenA=128°41'49", 


B= 107° 33' 20", 


c=124°12'31"; 




find a=125°44'44", 


b= 82° 47' 35", 


C=127°22' 7". 


54. 


Given A= 129° 58 '30", 


B= 34° 29' 30", 


c= 50° 6'20"; 




find a= 85° 59', 


b= 47° 29' 20", 


C = 36° 6' 50". 


55. 


Given A= 95° 38' 4", 


0= 97° 26' 29", 


b= 64° 23' 15"; 




find a= 99° 40' 48", 


c= 100° 49' 30", 


B= 65° 33' 10". 


56. 


Given A= 70°, 


B = 131°18', 


c=116°; 




find a= 57° 56' 53", 


5= 137° 20' 33", 


C= 94° 48' 12". 


57. 


Given a= 62° 15' 24", 


6=103°18'47", 


A= 53° 42' 38"; 




find B = 62°24'24".8, 


C = 155°43'll".3, 


c=153° 9' 351", 




or B'=117°35'35".2, 


C'= 59° 6' 10". 6, 


c'= 70° 25' 26". 


58. 


Given a= 52° 45' 20", 


b= 71°12'40", 


A= 46° 22' 10"; 




find B = 59°24'15f", 


C= 115° 39' 551", 


c= 97°33'18".8, 




or B'=120°35'44i", 


C'= 26°59'55".2, 


c'= 29°57'10".5. 


59. 


Given a= 48°45'40", 


6= 67°12'20", 


A= 42°20'30"; 




find B= 55° 39' 57", 


C= 116° 34' 18", 


c= 93° 8' 9".6, 




or B' = 124°20' 3", 


C'= 24° 32' 15", 


c'= 27° 37' 20". 


60. 


Given a= 46° 20' 45", 


b= 65° 18' 15", 


A= 40°10'30"; 




find B= 54° 6' 19", 


C = 116° 55' 26", 


c= 90°31'46", 




or B'=125°53'41", 


C'= 24°12'53".3, 


c'= 27° 23' 14". 



EXAMPLES. 



321 



Given a=150°57' 5", 
find B = 120° 47 '44", 
or B'= 59° 12' 16", 

Given a= 50°45'20", 
find B= 57° 34' 51 ".4, 
or B'=122°25' 8".6, 

Given a= 40° 5' 25". 6, 
find B= 42°37'17".5, 
or B'=137°22'42".5, 

Given a= 99°40'48", 
find B= 65° 33' 10", 
(No ambiguity ; why?) 

GivenA= 79°30'45", 

find 5= 36° 5'34f", 
(No ambiguity ; why?) 

Given A = 73° 11' 18", 

find b= 41°52'34f", 
(Only one solution ; why?) 

GivenA= 46°30'40", 
find 6= 33°18'47|", 
(Only one solution ; why?) 

Given A = 61° 29' 30", 

find b= 15°30'30".5, 
(Only one solution ; why?) 

Given A = 36° 20' 20", 
find b= 55° 25' 2|", 
or 6'=124°34'57|", 



b-- 
C= 

C'= 

b-- 
C-- 
C-- 

b-. 
G-- 
C'= 

b= 
C= 

B= 

c= 

B = 

c— 

B= 

c= 

B = 
c= 

B = 

c= 
c'= 



: 134° 15' 54", 

: 97° 42' 55", 
: 29° 9' 9", 

= 69° 12' 40", 
= 115° 57' 50". 6, 
= 25°44'31".6, 

=118°22' 7".3, 
=160° 1'24".4, 

= 50°18'55".2, 

= 64° 23' 15", 

= 97° 26' 29", 

= 46° 15' 15", 

= 50° 24' 57", 

= 61° 18' 12", 
= 41° 35' 4", 

= 36°20'20", 
-- 60° 32' 6", 

= 24°30'30", 
= 39° 33' 52", 



46° 30' 40", 

81° 27' 26 V, 
162° 34' 27", 



A= 

c'= 

A= 

c= 



c'= 



A-- 

c- 



: 144° 22' 42"; 
= 55° 42' 8", 
= 23° 57' 29". 

44° 22' 10"; 
95°18'16".4, 
28° 45' 5".2. 

29°42'33".8; 

153°38'42".4, 

90° 5'41".0. 

: 95° 38' 4"; 
= 100° 49' 30". 



a= 53°18'20"; 
C= 70° 55' 35". 

a= 46° 45' 30"; 
C = 60°42'46".5. 

o= 42°15'20"; 
C=110° 3'14".6. 



a= 34° 30"; 
C= 98°48'58".5. 



a= 42° 15' 20"; 

C=119°22'27£", 

C'=164°41'55". 



322 
70. 

71. 

72. 

73. 
74. 

75. 
76. 
77. 
78. 
79. 
80. 
81. 
82. 



SPHERICAL TRIGONOMETRY. 



Given A = 52° 50' 20", 
find 6= 81° 15' 15", 
or b'= 98° 44' 45", 

Given A= 115° 36' 45", 
find a=114°26'50", 

GivenA= 61°37'52".7, 
find a= 42°37'17".5, 
or a'=137°22'42".5, 

Given A = 70°, 
Ans. Impossible ; why ? 

Given a =108° 14', 
find A=123°53'47", 

Given a= 57° 17', 
find A= 21° V 2", 

Given a= 68° 45', 
find A= 94°52'40", 

Given a= 63° 54', 
find A= 86°30'40", 

Given a= 70° 14' 20", 
find A= 110° 51' 16", 

Given a=124°12'31", 
find A=127°22' 7", 

Given a== 50° 12' 4", 
find A= 59° 4'25", 

Given a=100°, 
find A=138°15'45".4, 

Given A = 86° 20', 
find a= 87° 20' 28", 



B= 


66° 7' 20", 


a= 


59° 28' 27"; 


c= 


110° 10' 50i", 


C= 


119° 43' 48", 


c'= 


138° 45' 26", 


C'» 


142° 24' 59". 


B = 


80° 19' 12", 


b= 


84°21'56"; 


c= 


82° 33' 31", 


C= 


79° 10' 30". 


B= 


139°54'34".4, 


b= 


150°17'26".2 


c= 


129° 41' 4".8, 


G= 


89°54'19".C 


c'= 


19°58'35".6, 


C'= 


26° 21' 17 ".C 


B= 


120°, 


b= 


80°. 


b= 


75° 29', 


c= 


56° 37'; 


B= 


57° 46' 56", 


C= 


46°51'51".£ 


b= 


20° 39', 


c= 


76° 22'; 


B = 


8°38'46", 


C = 


155° 31' 36'. c 


6= 


53° 15', 


c= 


46° 30'; 


B = 


58° 5' 10", 


C = 


50°50'52|". 


6= 


47° 18', 


c= 


53° 26'; 


B = 


54° 46' 14", 


C = 


63°12'55|" 


b= 


49° 24' 10", 


c= 


38° 46' 10"; 


B = 


48° 56' 4", 


C= 


38° 26' 48". 


6= 


54° 18' 16", 


c= 


97°12'25"; 


B= 


51° 18' 11", 


C = 


72° 26' 40". 


b= 


116° 44' 48", 


cm 


129° 11' 42"; 


B= 


94° 23' 10", 


C = 


120° 4' 50". 


b= 


50°, 


c= 


60°; 


B= 


31°11'14".0. 


C = 


35°49'58".L 


B = 


76° 30', 


C^ 


94° 40'; 


6= 


76°44'2f, 


e= 


93°55'31" 



EXAMPLES. 



323 



Given A = 96° 45', 
find a= 88°27'49", 

Given A = 78° 30', 
find a= 74° 57' 46", 

Given A = 57° 50', 
find a= 58° 8' 19", 

Given A =129° 5' 28", 
find a=135°49'20", 

Given A =138° 15' 50", 
find a=100° 0' 8".4, 

Given A =102° 14' 12", 
find a = 104°25' 8", 

Given A = 20° 9' 56", 
find a= 20° 16' 38", 



B = 108°30', 
6 = 107° 19' 52", 

B = 118°40', 
6=120° 8'49", 

B= 98° 20', 
b= 83° 5' 36", 

B = 142° 12' 42", 
6=144°37'15", 

B = 31° 11' 10", 
6= 49°59'56".4, 

B= 54° 32' 24", 
6== 53°49'25", 

B= 55° 52' 32", 
b= 56°19'41", 



C = 116°15'; 
c=115°28'13}". 
C= 93° 20'; 
c=100°18'll|". 

C= 63° 40'; 
c= 64° 3'20". 

C = 105° 8'10"; 
c= 60° 4' 54". 

C= 35° 50'; 
c= 60° 0'11".2. 

0= 89° 5'46"; 
c= 97° 44' 18". 

C = 114°20'14"; 
c= 66°20'43". 



90. If a, b, c are each < -, show that the greater angle may 
sceed -• 



| 91. If a alone >%■*, show that A must exceed -• 



•j 92. If a and b are each >|-7r, and c < \tt, prove that : 

(1) The greatest angle A must be >^ir ; 

(2) B may be > %■*; 

(3) C may or may not be < \tt. 



' 93. If cos a, cos b, cos c are all negative, prove that cos A, 
ts B, cos C are all necessarily negative. 

94. In a spherical triangle, of the five products, cosrt cos A, 
\s b cos B, cos c cos C, cos a cos b cos c, — cos A cos B cos C, show 
tat one is negative, the other four being positive. 



324 



SPHERICAL TRIGONOMETRY. 



CHAPTER XII. 



THE IN-OIEOLES AND EX-OIEOLES. — AEEAS. 



215. The In-Circle (Inscribed Circle).— To find the 
angular radius of the in-circle of a triangle. 

Let ABC be the triangle ; bisect the angles A and B 
by the arcs AO, BO ; from draw 
OD, OE, OF perpendicular to the 
sides. Then it may be shown that 
is the in-centre, and that the per- 
pendiculars OD, OE, OF are each 
equal to the required angular radius. 

Let 2 s = the sum of the sides of 
the triangle ABC. The right triangles 
OAE, OAF are equal. 

.-. AF = AE. 

Similarly, BD = BF, and CD = CE. 

.-. BC + AF = AC + BF = s. 
.-. AF = s-BC = s-a. 
Now tan OF = tan OAF sin AF . (Art. 186) 

or, denoting the radius OF by r, we have 




tan r = tan — sin (s - 

2 V 



a) 



(i) 



or tan r 



=4 



sin (s — a) sin (s — b) sin (s — c) 



sins 



n 
sins 



(Art. 195) (2) 



THE ESCRIBED CIRCLES. 325 

Also, sin(s— a) 



= sin 



*(&+«)-§ 



= smj(6 + c) cos-i-a — cos-J(& + c) sin|-a 
sin ^ a cos £ a 



. A 

sin — 

2 



[cos |(B - C) - cos i(B + C)] (Art. 198) 



- sing sin j-B sin^-C 

sin^-A 
which in (1) gives 

. B . C 

sm — sm — 

2 2 
tan r = —- — — sin a (3) 

cos|-A v y 

. . (Art. 196) (4) 



2 cos f A cos -J-B cos -J-C 
an equation which is equivalent to the following : 

cotr = — [cosS + cos(S-A)+cos(S-B)+cos(S-C)](5) 

u JN 

216. The Ex-Circles. — To find the angular radii of the 
ex-circles of a triangle. 

A circle which touches one side of a triangle and the 
other two sides produced, is called an escribed circle, or 
ex-circle, of the triangle. It is clear that the three ex-circles 
of any triangle are the in-circles of its colunar triangles 
(Art. 191, Sch.). 

Since the circle escribed to the side a of the triangle 
ABC is the in-circle of the colunar triangle A'BC, the parts 
of which are a, -n- — b, ir — c, 
A, 7r — B, 7r — C, the problem 
becomes identical with that 
of Art. 215 ; and we obtain 
the value for the in-radius of 
the colunar triangle A'BC, by substituting for 6, c, B, C, 
their supplements in the five equations of that article. 





326 SPHERICAL TRIGONOMETRY. 

Hence, denoting the radius by r a , we get 

tanr a = tan^-A sins (1) 

=>■ - (2) 

sm(s — a) 

cos^-Bcos^C • /ox 

= 2 —J — sina (3) 

cos^-A 

= £! (£\ 

2cos|AsiniBsiniC v ' 

cotr a =~— [-cosS-cos(S-A) + cos(S-B) + cos(S-C)](5) 

These formulae may also be found independently by 
methods similar to those employed in Art. 215, for the 
in-circle, as the student may show. 

Sell. Similarly, another triangle may be formed by pro- 
ducing BC, BA to meet again, and another by producing 
CA, CB to meet again. The colunar triangles on the sides 
b and c have each two parts, b and B, c and C, equal to 
parts of the primitive triangle, while their remaining parts 
are the supplements in the former case of a, c, A, C, and in 
the latter, of a, b, A, B. 

The values for the radii r b and r c are therefore found in 
the same way as the above values for r a \ or they may be 
obtained from the values of r a by advancing the letters. 

Thus, tan r b = tan^ B sin s = , etc., 

sm(s — b) 

and tan r c = tan 4 C sin s = — , etc. 

sm(s— c) 

217. The Circumcircle. — To find the angular radius of 
the eircumeirele of a triangle. 

The small circle passing through the vertices of a spheri- 
cal triangle is called the circumscribing circle, or circumcircle, 
of the triangle. 




THE CIRCUMCIRCLE. 327 

Let ABC be the triangle; bisect the 
sides CB, CA at D, E, and let be the 
intersection of perpendiculars to CB, 
CA, at D, E; then is the circum- 
centre. 

For ; join OA, OB, OC ; then (Art. 186) 

cos OB = cos BD cos OD, 
cos OC = cos DC cos OD. 
.-. OB = OC. Similarly, OC == OA. 
Now the angle 

OAB = OB A, OBC = OCB, OCA = OAC. 
.-. OCB + A=i(A + B + C)=S. 

... OCB = S-A. 
Let OC = E ; then, in the triangle ODC, we have 

cos OCD = tan CD cot CO = tan \ a cot R . (Art. 186) 

... tanE= tai ^ a A (1) 

cos(S-A) v ; 

or t anK = -~> (Art. 196) (2) 

Also cos (S - A) = cos £[ (B + C) - A] 

= cosi(B + C) cos|A + sin|-(B + C) sin|-A 

= Sm ^ Ac i OS ^ A [cos|(& + c) + cos^(6~c)] (Art. 198) 

COS ~2 Q> 

sin A 



cos^-a 



cos^fccos^c, 



which in (1) gives 



tanR = 5«L^ (3) 

sm A cosifrcos^-c 

= 2sini«sin^sin|c _ (Arfc 19g) (4) 



328 SPHERICAL TRIGONOMETRY. 

which may be reduced to the following : 

tan E = — [sin(s — a) + sin(s — b) + sin(s — c) — sin s] (5) 
2n 

218. Circumcircles of Colunar Triangles. — To find the 
angular radii of the circumcircles of the three colunar triangles. 

Let E 2 , E 2 , E 3 be the angular radii of the cireumcircles of 
the colunar triangles on the sides a, b, c, respectively. 
Then, since E x is the eircumradius of the triangle A'BC 
whose parts are a, w — b, ir — c, A, ir — B, ir — C, we have, 
from Art. 217, 

tanE 1 = -^5i^ (1) 

cosS 



tan 



Ei = cos(S-A) (2) 



N 



tanE 1 = -, ^i^ (3) 

sin Asm-J-6 sm-J-c 

tanE = 2 sin % a cos ^ b cos ^ c (4) 

n 

tanEj = — [sins— sin(s— a) + sin(s— &)+sin(s— c)](5) 

2n 

Similarly, 

tanR 2 = -^M = ^(S-B) =et 

cos S 1ST 

and tanR 3 = -^ i ^ = C ° S(S ~ C) =etc. 
cosS N 

EXAMPLES. 

Prove the following : 

1. cos s + cos(s — a) + cos(s — b) + cos(s — c) 

=s 4 cos £ a cos £ 6 cos £ c. 

2. cos (s — b) + cos (s — c) — cos (s — a) — cos s 

= 4 cos |- a sin £ 6 sin -J- a 



PROBLEM. 329 



N 



3. tann= cosiCcosiA sin& = 

cosf JB 2cosfi3sm^-C sm-J-A 

. cos -t A cos ^-B . N 

4. tan r c = — sm c = • 

cos-J-C 2cos^-C sin^Asin|-B 

5. cot r : cot i\ : cot r 2 : cot r 3 

= sins : sin(s — a) : sin(s — b) : sin(s — c). 

6. tan r tan i\ tan r 2 tan r s = ?i 2 . 

7. cot r tan r x tan ?* 2 tan p 3 = sin 2 s. 
£ t "R — 2 cos |- a sin ^ b cos j- c 

9 tanB ^ 2 CQS i a CQS l&sinj-c 

10. tanE 1 :tanE 2 :tanE 3 =cos(S — A):cos(S— B):cos(S — C). 

11. cot E cot E x cot E 2 cot E 3 = N 2 . 

12. tan E cot E x cot E 2 cot E 3 = cos 2 S. 

AREAS OF TRIANGLES. 

219. Problem. — To find the area of a spherical triangle, 
having given the three angles. 

Let r = the radius of the sphere. 

E = the spherical excess = A + B + C - 180°. 
K = area of triangle ABC. 

It is shown in Geometry (Art. 738) that the absolute area 
of a spherical triangle is to that of the surface of the sphere 
as its spherical excess, in degrees, is to 720°. 

.-. K:47rr 2 = E:720°. 

... K^-^-Trr 2 (1) 

180° v J 



330 SPHERICAL TRIGONOMETRY. 

Cor. The areas of the colunar triangles are 

(2A-E) 2 (2B-E) . (2C-E) ^ 
180° ' 180° ' 180° 

220. Problem. — To find the area of a triangle, having 
given the three sides. 

Here the object is to express E in terms of the sides. 

I. Cagnolfs Theorem. 

sin £E = sin | ( A + B + C - tt) 

= sin|(A + B) sin ±C — cos £(A + B) cos £C 

= smiCj^siC [ cos i(a-6)~cos|(a + 6)] . (Art. 198) 
sin J a sin -J- 6 sin C __ sin -J- a sin -J- 6 2n /A , 1QKUn 

•— — ; — ; ' —. : — 7 V Art - -L^O J ^1 J 

cos f c cos -^ c sin a sin o 

.% sin£E = ^- — (2) 

2cosf a cos -J- b cos-J-c 

II. Lhuilier's Theorem. 

sini(A + B + C-^ 
4 cosi(A + B + C-7r) 

_ sin |(A + B) - sin j-(V - C) ^ Arf 4g , 

cos£(A + B) + cos£(jt-C) V ' 

_ sin|(A + B) — cos^-C 
~~ cos |(A + B) + sin £C 

_ cos|-(a-6)-cos|c _ cos|C ^ , Aj>t 19g , 

cos£(a + 6) + cos^-c sin^C 

= sin|( g -6) S in^-a) cotiC! (Art. 45) 

cos-^-s cos£(s — c) 

= Vtan -J- stands — a) tan|(s — &)tan-J-(s— c) (Art. 195) (3) 



ABE AS OF TRIANGLES. 



331 



221. Problem. — To find the area of a triangle, having 
given two sides and the included angle. 

cosiE=cos[|(A + B)-(i7r-iC)] 

= cos£(A + B) sin£C + sin£(A + B) cos^C 

= cosi(a + 6)sm 2 |C + cos^-(a-6)cos 2 |-C (Art. 198) 

= [cos|-acos|-& + sin|asin^&cosC]sec|-c . . (1) 

Dividing (1) of Art. 220 by this equation, and reducing, 
we have 

tanjE= tanjatanpsinC (2) 

1 + tan 4- a tan+ 6 cos C 



EXAMPLES. 

1. Given a==H3°2'56".64, b = 82°39'28".4, c = 74°54' 
31 ".06; find the area of the triangle, the radius of the 
sphere being r. 

By formula (3) of Art. 220, 

a = 113° 2'56".64 



6= 82°39'28".40 
c= 74° 54' 31 ".06 

2* = 270°36'56".10 



s = 135°18'28".05 
s-a= 22° 15' 31 ".41, 
s-6= 52°38'59".65, 
s-c= 60°23'56".99. 



K: 



2 9 3 2 7 3 
648000 



Js = 67°39'14".025 
£(s-a) = ll° 7' 45". 705 
£(s-&) = 26°19'29".825 
£(s-c) = 30°ll'58".495 

log tan ^s = 0.3860840 
log tan i (s - a) = 9.2938583 
log tan i(s -b)= 9.6944058 
log tan %(s-c) = 9.7649261 

logtan 2 iE = 9.1392742 
log tan £E = 9.5696371. 

1E = 20°21'58".25. 
E = 81° 27' 53" 
= 293273". 
XTrr 2 . . . . [(1) of Art. 219] 



_ 2 9 3 2 7 3 ,.2 
— TTT62 



f = area. 



332 SPHERICAL TRIGONOMETRY. 

2. Given A = 84° 20' 19", B = 27°22'40", C = 75°33'; 
find E = 7°15'59". 

3. Given a = 46° 24', b = 67° 14', c = 81° 12' ; 

fi^d TT _ JJ..8 3 8.0. r 2 

I1I1U XS- — 206265' • 

4. Given a = 108° 14', b = 75° 29', c = 56° 37' ; 
find E = 48° 32' 34".5. 



5. Prove cos^E = 1 + cosa + cos& + cosc 
4cos|-acosi& cos^c 

cos 2 ia + cos 2 i& + cos 2 |c — 1 



2cos^-a cos^-6 cos|-c 



6. " sin 1 E=r V p n ^ gsin "^ s ~ a ^ sin ^^~^ sin ^^'~^ 
A/ cos^-a cos^fr cos+c. 



rt" IAJ KjVJiJ IT W (jUOyl 



7. " cos 1 E= JGQ^sGosi(s-a)cos^{s-b)G Os\(8-c) 
\ cos + acoslfr cosic 



8 " cotiE = cot|acot^5 + cosC 
2 sinC 

_ cot^b cot^-c + cos A 



sin A 
_ coticcot^-a + cosB ^ 



sinB 
EXAMPLES. 

Prove the following : 

1. sin(s — a) + sin(s — b) + sin(s •— c) — sins 

= 4 sin-i-a sinj& sin|-c. 

2. sins + sin(s — b) + sin(s — c) — sin(s — a) 

= 4sin|-a cos^-6 cos-J-c. 



EXAMPLES. 333 

3. sin (s — b) sin (s — c) + sin (s — c) sin (s — a) 

+ sin (s — a) sin (s — 6) + sin s sin (s — a) 
+ sins sin(s — b) + sins sin(s — c) 
== sin 6 sin c + sin c sin a + sin a sin b. 

4. sin (s — b) sin (s — c) + sin (s — c) sin (s — a) 

— sin(s — a)sin(s — b) + sins sin(s— -a) 
+ sins sin(s — b) — sins sin(s — c) 
= sin 6 sinc + sincsina— sinasinfr. 

5. sin 2 s + sin 2 (s — a) + sin 2 (s — 6) + sin 2 (s — c) 

= 2(1 — cosacos&cosc). 

6. sin 2 s + sin 2 (s — a) — sin 2 (s — b) — sin 2 (s — c) 

= 2 cos a sin b sine. 

7. cos 2 s + cos 2 (s — a) + cos 2 (s — 6) + cos 2 (s — c) 

= 2(1 + cos a cos 6 cose). 

8. cos 2 s + cos 2 (s — a) — cos 2 (s — b) — cos 2 (s — c) 

= — 2 cos a sin b sin c. 

9. tan r cot r x tan r 2 tan r 3 = sin 2 (s — a) . 

10. tan?* tanr x cotr 2 tanr 3 = sin 2 (s — 6). 

11. tan r tan 7\ tan r 2 cot r 3 = sin 2 (s — c) . 

12. cot r sins = cot-|-A cot-|-B cot^-C. 

4]STsinS 



13. tan r x + tan r 2 + tan r 3 — tan r = 

sin A sin B sin C 

14. cotr 1 + cotr 2 + cotr 8 -cotr = 4sin * asi ^ &sin ^ c - 

15. tanr i: tanr 2 = tanr,= sin(t : sin b : sln c ■ 

1 + cos A 1 + cos B 1 + cos C 

m tan?%+tanr 2 +tanro— tan r , ,+ , , , , v 

16. — — ^ — ^ s = i(l+cosa + cos6 + cosc). 

cot r x + cot r 2 + cot r 3 — cot r 



334 



SPHERICAL TRIG ONOMETR Y. 



17. 

18. 
19. 



cot^ + cotfr, + cot«r, + cot'r = 2(l-cosaco86cosc) . 



1 +■ * 



2 cos a sin 6 sin c 



snr'r sm 2 ^ sin 2 r 2 sin 2 r 3 



cot r 2 cot r 3 + cot r s cot r x + cot o\ cot r 2 

+ cot r (cot r x + cot r 2 + cot r 3 ) 
_ sin b sin c + sin c sin a + sin a sin b 



20. 



21. 

22. 
23, 
24. 
25. 
26. 
27. 
28. 
29. 
30. 
31. 
32. 

33. 
34. 



tan r 2 tan r s + tan r 3 tan 7\ + tan r x tan r 2 

+ tan r (tan r x + tan r 2 + tan r 3 ) 
= sin 6 sin c + sin c sin a + sin a sin 6. 

cot E tan B x cot E 2 cot E 3 = cos 2 (S — A) . 
cot E cot E x tan E 2 cot E 3 = cos 2 (S — B). 
cot E cot E x cot E 2 tan E 3 = cos 2 (S — C). 
tan E x + tan E 2 = cot r + cot r 3 . 
tan E x + tan E 2 + tan E 3 — tan E = 2 cot r. 
tan E — tan E 2 + tan E 2 + tan E 3 = 2 cot r v 
tan E + tan E x — tan E 2 + tan E 3 = 2 cot r 2 . 
tan E + tan Ei + tan E 2 — tan E 3 = 2 cot r 3 . 
cot r 2 + cot r 2 + cot r z — cot r = 2 tan E. 
cot r — cot r x + cot r 2 + cot r 3 = 2 tan E x . 
cot r + cot r x — cot r 2 + cot r 3 = 2 tan E 2 . 
cot r + cot r 2 + cot r 2 — cot r 3 = 2 tan E 3 . 

tan E + cot r = tan E x + cot r 2 = etc., 

= |(cot r 4- cot 7 1 ! + cot r 2 + cot r 3 ). 

tan 2 E + tan 2 E x + tan 2 E 2 + tan 2 E 3 

- 2(1+ cos A cos B cos C) 

N 2 



EXAMPLES. 335 

ok tan 2 B + tan 2 Bj + tan 2 E 2 + tan 2 B 3 _ ^ 
cot 2 r + cot 2 rj + cot 2 r 2 + cot 2 r z 

36. tan 2 E + tan 2 E x - tan 2 E 2 - tan 2 E 3 

2 (cos A sinB sinC) 

N 2 

37. cot 2 r + cot 2 r 1 -cot 2 r 2 -cot 2 r 3 = ^^^ A ^- C - 

re 

on tan 2 E + tan 2 Ex — tan 2 E 2 — tan 2 E 3 _ __ cos A 
cot 2 r + cot 2 r x — cot 2 r 2 — cot 2 r 3 cos a 

39. tan E cot E 2 = tan -J- 6 tan -^ c. 

.a , , , , -DX9.-I /sin a + sin b + sin c\ 2 

40. (cot r + tan E) 2 + 1 = — — ) • 

V 2n J 

41. (cot r, - tan E) 2 + 1 = f sin b + s ^ c " sin a Y- 

N 



42. tan ^ A sin (s — a) ■■ 



2 cos % A cos ^B cos £C 



4S tan?- _ cos(S — A) cos(S — B) cos(S — C) 
tanE 2cos^-Acos-J-Bcos-J-C 

44. cot (s — b) cot (s — c) + cot (s — c) cot (s — a) 

+ cot(s — a) cot(s — b) = cosec 2 r. 

45. eot (s — 6) cot (s — c) — cot s cot (s — 6) — cot s cot (s — c) 

= cosec 2 ?y 

46. cot(s— c)cot(s— a) — cotscot(s— c) — cotscot(s— a) 

= cosec 2 r 2 . 

47. cot (s — a) cot (s—b) — cot s cot (s — a) — cot s cot (s — b) 

== cosec 2 ?* 3 . 

,g cot (s — a) cot(s — b) cot(s — c) 2 cot s 
sin 2 t*! sin 2 r 2 sin 2 r 3 sin 2 r 

= 3cot(s — a)cot(s — b)cot(s — c). 



336 SPHERICAL TRIGONOMETRY. 



49. cosec 2 ^ + cosec 2 r 2 + cosec 2 r 3 — cosec 2 r 






= — 2 cot s[cot(s — a) + cot(s — b) + cot(s — c)]. 



50. -±- + -±- + -J- + 



snrr sm-?*! sm 2 r 2 sin 2 r 3 



_ — 2Stan(s — g)tan(s — b) 

tan s tan(s — a) tan (s — b) tan(s — c) 

51. cot E -* cot B 2 - cot E 2 - cot E 3 = 2 X cos s . 

n 

52. sin(A-£E) = - 

2 cos $ a sin £ 6 sin £ c 

53. sin(B-£E) = - 

2 sm ^a cos -J- b sin ^-c 

54. sin(C-iE) = ^ - 

2 sm-^-a sin £6 cos £c 

trr /a i -i7>\ sin 2 £& + sin 2 ^-c — sin 2 £a 

55. cos(A — -J-E) = * — X 2 ? — 

2 cos £ a sin -J- 6 sin-J-c 

m /-d i tt«\ sin 2 Ac + sin 2 4a — sin 2 -J- 6 

56. cos(B— ^E) = 2 — ■ * 2 — 

2sm- 2 -acos£fr sm^-c 

eir //-< ip\ sin 2 -J- a + siir^fr — sin 2 Ac 

57. cos(C — £E) = — „ T . 7 — ^V; r- 2 -' 

2 sin-^-a sm -jo cos fc 

58. cot(A-|E) = cot * at *"P- cos C 

sin C 

_ tan -J- 6 tan £ c + cos A _ tan -J- a cot ?b — cos B 
sin A . sin B 



59. tan-^(A — iE) = Vcot|scoti(s— a)tani(s — 6)tan^-(s— c). 



60. tani(B — £E) = VcotJstan£(s— a)cot|(s— 6)tan£(s— c). 



61. tani(C— ^E)=Vcot^stan^(s— a)tan|(s— 6)cot|(«— c). 

62. If Sj S b So, S 8 denote the sums of the angles of a triangle 
and its three colunars, prove that S + Si + S 2 + S 3 = 3 w. 

p3. Tn an equilateral triangle, tan E = 2 tan?*. 



EXAMPLES. 337 

64. If E 1? E 2 , E 3 denote the spherical excesses of the colunars 
on a, b, c, respectively, show that E + E x + E 2 + E 3 = 2ir ; 
and therefore the sum of the areas of any triangle and its 
colunars is half the area of the sphere. 

65. Given a = 108° 14', b = 75° 29', c = 56°37'; 

find E = 48° 32' 34".5. 

66. Given a = 63° 54', 6 = 47° 18', c = 53°26'; 

find E = 24° 29' 491". 

67. Given a = 69° 15' 6", b = 120° 42' 47", c = 159° 18' 33"; 

find E = 216° 40' 23". 

68. Given a = 33°l'45", b = 155° 5' 18", C = 110°10'; 

find E = 133° 48' 55". 

69. Given a = b = c = 1°, on the earth's surface ; 

find E = 27".21. 

70. Given a = b = c = 60°, on a sphere of 6 inches radius ; 
find the area of the triangle. Ans. 19.845 square inches. 

71. If a=b = -, and c=— } prove sin-£E = i, and cosE=|-. 

72. If C = ~ f prove sin -i- E = sin \ a sin \ b sec \ c, and 
cos|-E = cosiacos^6 sec|-c. 

73. If a = b, and C = -, prove tan E = \ tan a sec a. 

74. If A+B + C = 2tt, prove cos 2 A- a + cos 2 1 6 + cos 2 J-c=l. 

75. If a + b = 7r, prove that E = C ; and if E' denote the 
spherical excess of the polar triangle, prove that 

sin^-E'= sin a cos ^C. 



76. Prove sinUE = V sin * E si " i E - sinjE 2 sin£E 8j 

cot i a cot %b cot £c 



338 SPHERICAL TRIGONOMETRY. 



CHAPTER XIII. 
APPLICATIONS OP SPHEKIOAL TKIGONOMETKY. 

SPHERICAL ASTRONOMY. 

222. Astronomical Definitions. 

The celestial sphere is the imaginary concave surface of 
the visible heavens in which all the heavenly bodies appear 
to be situated. 

The sensible horizon of a place is the circle in which a 
plane tangent to the earth's surface at the place meets 
the celestial sphere. 

The rational horizon is the great circle in which a plane 
through the centre of the earth parallel to the sensible 
horizon meets the celestial sphere. Because the radius 
of the celestial sphere is so great, in comparison with the 
radius of the earth, these two horizons will sensibly 
coincide, and form a great circle called the celestial hori- 
zon. 

The zenith of a place is that pole of the horizon which 
is exactly overhead ; the other pole of the horizon directly 
underneath is called the nadir. 

Vertical circles are great circles passing through the 
zenith and nadir. The two principal vertical circles are 
the celestial meridian and the prime vertical. 

The celestial meridian of a place is the great circle in 
which the plane of the terrestrial meridian meets the 
celestial sphere ; the points in which it cuts the horizon 
are called the north and south points. 



SPHERICAL ASTRONOMY. 339 

The prime vertical is the vertical circle which is per- 
pendicular to the meridian ; the points in which it cuts 
the horizon are called the east and ivest points. 

The axis of the earth or of the celestial sphere is the 
imaginary line about which the earth rotates. 

The celestial equator, or equinoctial, is the great circle in 
which the plane of the earth's equator intersects the 
celestial sphere. 

The poles of the equinoctial are the points in which the 
axis pierces the celestial sphere. 

Hour circles, or circles of declination, are great circles 
passing through the poles of the equinoctial. 

The ecliptic is a great circle of the celestial sphere, and 
the apparent path of the sun due to the real motion of 
the earth round the sun. 

The equinoxes are the points in which the ecliptic cuts 
the equinoctial. There are two, called the vernal and the 
autumnal equinox, which the sun passes on March 20 and 
September 22. 

The obliquity of the ecliptic is the angle between the 
planes of the ecliptic and equator, and is about 23° 27'. 

Circles of latitude are great circles passing through the 
poles of the ecliptic. 

223. Spherical Coordinates. — The position of a point 
on the celestial sphere may be denoted by any one of three 
systems. In each system two great circles are taken as 
standards of reference, and the point is determined by 
means of these circles, which are called its spherical 
coordinates, as follows : 

I. The horizon and the celesticd meridian of the place. 

The azimuth of a star is the arc of the horizon inter- 
cepted between the south point and the vertical circle 



340 SPHERICAL TRIGONOMETRY. 

passing through the star; it is generally reckoned from 
the south point of the horizon round by the west, from 
0° to 360°. 

The altitude of a star is its angular distance above the 
horizon, measured on a vertical circle. The complement 
of the altitude is called the zenith distance. 

II. The equinoctial and the hour circle through the vernal 
equinox. 

The right ascension of 'a star is the arc of the equinoctial 
included between the vernal equinox and the hour circle 
passing through the star; it is reckoned eastward from 
0° to 360°, or from h to 24\ 

The angle at the pole between the hour circle of the 
star and the meridian of the place is called the hour angle 
of the star. 

The declination of a star is its distance from the equinoc- 
tial, measured on its hour circle ; it may be north or south, 
and is usually reckoned from 0° to 90°. It corresponds to 
terrestrial latitude. 

The polar distance of a star is its distance from the pole, 
and is the complement of its declination. The right ascen- 
sion and declination of celestial bodies are given in nautical 
alrnxtnacs. 

III. The ecliptic and the circle of latitude through the vernal 
equinox. 

The latitude of a star is its angular distance from the 
ecliptic measured on a circle of latitude ; it may be north 
or south, and is reckoned from 0° to 90°. 

The longitude of a star is the arc of the ecliptic inter- 
cepted between the vernal equinox and the circle of latitude 
passing through the star. 



SPHERICAL COORDINATES. 



341 



224. Graphic Representation of the Spherical Coordi- 
nates. — The figure will 
serve to illustrate the pre- 
ceding definitions. is 
the earth, PHP'R is the 
meridian, P the north pole, 
HR the horizon, EQ the 
equinoctial, Z the zenith. 
Then, of a place whose 
zenith is Z, QZ is the ter- 
restrial latitude ; and since 

QZ = PE, 

.*. PR = the latitude. 

But PR is the elevation of the pole above the horizon. 

Hence the elevation of the pole above the horizon is equal to 
the latitude. 

Let V be the vernal equinox, and let S be any heavenly- 
body, such as the sun or a star ; then its position is denoted 
as follows : 




VK = right ascension of the 


body 


BO, 


KS = declination « " 


a 


= 8, 


ZPS or QK = hour angle " " 


a 


= t, 


PS = north polar distance " " 


a 


=P, 


HT = azimuth " " 


it 


= a, 


TS = altitude " " 


u 


=h, 


ZS = zenith distance " " 


a 


= z, 


QZ = PR = latitude of the obsers 


r er 


= +. 



The triangle ZPS is called the astronomical triangle; 
ZP = 90° — <f> = co-latitude of the observer, 



PS = 90° -8, SZ = 90°-A. 



342 



SPHERICAL TRIGONOMETRY. 



Let the small circle MM', passing through S, and parallel 
to the equinoctial, represent the apparent diurnal motion of 
the heavenly body S (the declination being supposed con- 
stant); then the body S will appear to rise at A (if we sup- 
pose the Eastern hemisphere is represented in the diagram). 
It will be at B at 6 o'clock in the morning, at M at noon, 
at M' at midnight, and at <o it will be east. 

225. Problems. — By means of the foregoing definitions 
and diagram we may solve several astronomical problems of 
an elementary character as follows : 

(1) Given the latitude of a place and the declination of a 
+tar ; to find the time of its rising. 

Let A be the position of the star in the horizon. Then 
in the triangle APE, right angled at E, we have 

cos EPA = - cos ZPA = tan EP cot AP. 

.-. cos t = — tan <£tanS (1) 

from which the hour angle is found. 

Since the hourly rate at which a heavenly body appears 
to move from east to west is 15°, if the hour angle be 
divided by 15 the time will be found. In the case of the 
sun, formula (1) gives the time from sunrise to noon, and 
hence the length of the day. 

Ex. Eequired the apparent time of sunrise at a place 
whose latitude is 40° 36' 23".9, on July 4, 1881, when the 
sun's declination is 22° 52' 1". 



</> = 40° 36' 23".9, 
8 = 22° 52' 1". 



log tan <£ = 9.9331352 
log tan 8 = 9.6250362 

log cos t 



9.5581714- 
t = 111° 11' 44" 
= j* 24 m 47 8 , nearly, 






PROBLEMS OF SPHERICAL ASTRONOMY. 343 

which taken from 12 h , the time of apparent noon, gives 
4 h 35 m 13 s , the time of apparent sunrise.* 

(2) Given the latitude of a place and the declination of a 
star; to find its azimuth from the north at rising. 

Let A = the azimuth = AK. Then in the triangle APE 
we have 

cos AP = cos AE cos PE, 

or sin 8 = cos A cos cj>. 

.-. cos A= sin 8 sec <f> (2) 

Ex. Eequired the hour angle and azimuth of Arcturus 
when it rises to an observer in New York, lat. 40° 42' N., 
the declination being 19° 57' 1ST. 

Ans. 7 h 12 m 46 s .3 ; K 63° 15' 11" E. 

(3) Given the latitude of the observer and the hour angle 
and declination of a star ; to find its azimuth and altitude. 

Here we have given, in the triangle ZPS, two sides and 
the included angle; that is, PZ = 90° - <f>, PS = 90° - 8, 
and ZPS = t. Let A = the azimuth from the north = ET, 
p = the angle ZSP, and z = ZS. Then by Delambre's 
Analogies (Art. 198), 

sin|(p + A) cos-|-2= cos|-(S — <£) cos-|-£, 

sin^(p — A) sin^z = sin|-(8 — <£)cos|-£, 

cos ^(p + A) cos \ z = sin ^(8 + <£) sin % t, 

cos|-(p — A) sin \ z = cos|(8+ <£) sin-^. 

Hence, when <j>, t, and 8 are given, that is, the latitude of 
the place, and the hour angle and the declination of a 
heavenly body, A, z, and p can be found. 

In a similar manner may be solved the converse problem : 
Given the latitude of the observer and the azimuth and altitude 
of a star ; to find its hour angle and, declination. 

* In these examples no corrections are applied for refraction, semi-diameter ol 
the sun, change in declination from noon, etc. 



344 



SPHERICAL TRIGONOMETRY. 



(4) Given the right ascensions and declinations of two stars; 
to find the distance between them. 

Let P be the pole, S and S' the two stars. p a _ a r 
Let a and a' be the right ascensions of the 
stars ; 

S and 8' their declinations ; and 

d the required distance. 

Then we have given, in the triangle PSS', 
two sides and the included angle ; that is, 
PS = 90° - 8 = p, PS' = 90° - 8' =p\ and 
P = a - a'. 

This may be solved by Art. 198, or by the 
second method of Art. 209, as follows : 

Draw SD perpendicular to PS' produced; let PD = m. 
Then 

cosP = tanPDcotPS. 




.\ tan m = cos P tan p. 
Also cos SS' = cos PS cos S'D sec PD 
.\ cos d = cosp cos S'D sec m. 



(Art. 209) 



Ex. Eequired the distance between Sirius and Aldebaran, 
the right ascensions being 6 h 38 m 37 8 .6 and 4 h 27 m 25 8 .9, and 
the declinations 16° 31' 2" S. and 16° 12' 27" K, respectively. 



Here P = 2 h ll m 11 8 .7 

= 32° 47' 55" 

# = 106° 31' 2" 

m = 109° 25' 55" 

p'= 73° 47' 33" 

S'D= 35° 38' 22" 

jp = 106°31' 2", 

m = 109° 25' 55", 

SS' = d= 46° 0'44". 



log cos P = 9.9245789 
log tanp = 05279161 - 
log tan m = 0.4524950- 

109° 25' 55". 



m = 



log cos S'D = 9.9099302 
log cosp = 9.4537823- 

colog cos m = 0.4779643 - 
log cos d = 9.8416768 



PROBLEMS OF SPHERICAL ASTRONOMY. 345 



(5) Given the right ascension and declination of a star ; to 
find its latitude and longitude. 

Let V be the vernal equinox, S the star, VD, VL the 
the equator and the ecliptic, SD, SL 
perpendicular to VD, VL. Then 
VD = right ascension = a, SD = dec- 
lination = 8, VL == longitude = A, 
SL = latitude = I. Denote the ob- 
liquity of the ecliptic DVL by <o, 
and the angle DVS by 6. 

From the right triangles SVD, 
SVL we get 

cot 6 = sin a cot S . (1) 

tanA = cos (0 — <d) tana sec 6 (2) 

sinZ= sin (0 — co) sin 8 cosec0 (3) 

From (1), is determined; and from (2) and (3), X and 
I are determined. 

Ex. Given the right ascension of a star 5 h 6 m 42 s .01, and 
its declination 45° 51' 20".l N.; to find its longitude and 
latitude, the obliquity of the ecliptic being 23° 27' 19".45. 




a = 76° 40' 30".15 
8 = 45° 51' 20".l 
= 46° 38' 11 ".8 
a) == 23° 27' 19". 45 

0-o> = 23°lO'52".35 
A = 79° 58' 3".44. 



Z = 22°51'48".4. 



log sin a = 9.9881479 
log cot 8 = 9.9870277 

log cot = 9.9751756 

log cos (0 - o>) = 9.9634401 

log tan a = 10.6255266 

cologcos0 = 0.1632816 

log tanX = 10.7522483 

log sin (0 - o>) = 9.5950996 

log sin 8 = 9.8558743 

cologsinfl= 0.1384575 

log sin I = 9.5894314 



346 SPHERICAL TRIGONOMETRY. 



EXAMPLES. 

1. Find the apparent time of sunrise at a place whose 
latitude is 40° 42', when the sun's declination is 17° 49' N. 

Ans. 4 h 56 m . 

2. Given the latitude of a place = 40° 36' 23".9, the hour 
angle of a star=46°40'4".5, and its declination = 23° 4' 24".3 ; 
to find its azimuth and altitude. 

Ans. Azimuth = 80° 23' 4 ".47, altitude = 47° 15' 18".3. 

3. Find the altitude and azimuth of a star to an observer 
in latitude 38° 53' N"., when the hour angle of the star is 
3 k 15 m 20 s W., and the declination is 12° 42' N. 

Ans. Altitude = 39° 38' 0"; azimuth = S. 72° 28' 14" W. 

4. Given the latitudes of New York City and Liverpool 
40° 42' 44" K and 53° 25' N., respectively, and their longi- 
tudes 74°0'24"W. and 3° W., respectively; to find the 
shortest distance on the earth's surface between them in 
miles, considering the earth as a perfect sphere whose 
radius is 3956 miles. 

Note. — This is evidently a case of (4) where two sides and the included angle 
are given, to find the third side. 

Ans. 3305 miles. 

5. The latitudes of Paris and Pekin are 48° 50' 14" K 
and 39° 54' 13" 1ST., and their difference of longitude is 
114° 7' 30"; find the distance between them in degrees. 

Ans. 73° 56' 40". 

GEODESY. 

226. The Chord al Triangle. — Given two sides and the 
included angle of a spherical triangle; to find the correspond- 
ing angle of the chordal triangle. 



GEODESY. 



347 




The chordal triangle is the triangle formed 
by the chords of the sides of a spherical 
triangle. 

Let ABC be a spherical triangle, the 
centre of the sphere, A'BC the colnnar 
triangle, and M, N". the middle points of 
the arcs A'B, A'C. Then the chord AB is 
parallel to the radius OM, since they are 
both perpendicular to the chord A'B. Simi- 
larly, AC is parallel to ON. 

In the spherical* triangle A'MN", we have 

cos MN = cos A'N cos A'M + sin A'N sin A'M cos A' (Art. 191) 

Denote the angle BAC of the chordal triangle by A x . 
Then arc MM" or angle MOIST = A 1? A'N = J(w - b), 
A'M = 1(tt-c), and A' = A. 

.*. cos A 1= = sin -J- 6 sin|-c + cos^-6 cos|-ccos A . (1) 
with similar valnes for cos B x and cos C^ 

Cor. 1. If the sides b and c are small compared with 
the radius of the sphere, A 1 will not differ much from A. 

Let Aj = A — ; then 

cos A x = cos A + sin A, nearly. 

But sin-J-6 sin-Jc = sin 2 i(6 + c) — sin 2 ^(6 — c), 

and cos^b cos|-c = cos 2 i(6 + c) — sin 2 i(& — c). 

Substituting in (1) and reducing, we get 

(9 = tan-i-Asin 2 i(& + c)-coti-Asin 2 -i(&-c) . (2) 

which is the circular measure of the excess of an angle of the 
spherical triangle over the corresponding angle of the chordal 
triangle. 

The value in seconds is obtained by dividing the circular 
measure by the circular measure of one second, or, approxi- 
mately, by the sine of one second. 



348 SPHERICAL TRIGONOMETRY. 

Cor. 2. The angles of the chordal triangle are, respectively, 
equal to the arcs joining the middle points of the sides of the 
colunar triangles. 

227. Legendre's Theorem. — If the sides of a spherical 
triangle be small compared ivith the radius of the sphere, then 
each angle of the spherical triangle exceeds by one-third of the 
spherical excess the corresponding angle of the plane triangle, 
the sides of which are of the same length as the arcs of the 
spherical triangle. 

Let a, b, c be the lengths of the sides of the spherical 
triangle, and r the radius of the sphere ; then the circular 

measures * of the sides are respectively -, -, — Hence, 

r r r 

neglecting powers of — above the fourth, 
r 



cos 


a 


b c 

- cos -COS - 










cos A = — 


T 

sir 


r 

b . c 

i-sin- 

r r 


r 








(Art. 191) 


(l *4 


24ry V 


62 + 
2c 2 


&4 Yi- 


c 2 

2r 2 


24rV 


(Art. 156) 






^- 


6r»A 








_rY& 2 + , 


,a __ 


a 2 a 4 


-6 4 -, 


c 4 — 6 b 2 c 


■X 1 - 


b 2 +c 
Or* 


•V 1 


~bc\ 2r> 


1 


24 


r 4 


J 


_/6 2 + c 2 - 


-a 2 


i ai - 


& 4 -c 4 - 


-66WV 


I+J 


• 2 + <A 
Gj- 2 y 




i, 26c 




1 


246cr» 


A 




b 2 + c>- 


a 2 


26 2 c 2 + 2c 2 a 2 


+ 2a 2 6 2 - 


-a 4 - 


- b 4 - e 


! . . . a\ 



26c 24 bcr 2 

* The terra - is the circular measure of the angle which the arc a subtends at the 

r be 

centre of the sphere ; and similarly for - and — 



GEODESY. 349 

Now if A', B' ? C denote the angles of the plane triangle 
whose sides are a, b, c, respectively, we have 

cosA' = k 2 + c ' 2 ~ a2 (Art. 96) 

2 be 

, . BAI 2b 2 c 2 + 2c 2 a 2 + 2a 2 b 2 -a*-b 4 -c 4 /A . 1AA . 

and snrA' = — ' (Art. 100) 

4roV 

Therefore (1) becomes 

A A , &csin 2 A' /0 v 

cos A = cos A' .... (2) 

Let A = A r 4- 0, where is a very small quantity ; then 
cos A = cos A' — sin A f , nearly. 

g = 6csinA' = A (Art. 101) 

6^ 3r 2 V y 

where A denotes the area of the plane triangle whose sides 
are a, b, c. 

We have therefore A = A f -\ 

3r* 

Similarly B = B' + A, G == C + A. 
3 r 6 ir 

... A + B + C - A' - B' - C = - t |; 

r 2 

or A + B + C — 7T = — = spherical excess (Art. 219) 



_A_ 
3r* 



A — A'=B — B'=C — C f = ^3= i spherical excess. 



Cor. 1. If the sides of a spherical triangle be very small 
compared with the radius of the sphere, the area of the 
spherical triangle is approximately equal to 

*{l + 



\ 24 r 2 J 



350 



SPHERICAL TRIGONOMETRY. 



For, taniE= A /tan — tan^^tan— tan — (Art. 220^ 
4 \ 2r 2r 2r 2r V 



and 
tan 



2r 2r\_ 



12 1* 



; tan 



?— a s— a 



2r 



2r 



1 + 



(s-a) 2 
"l2r» 



, etc. 



(Art. 156) 



: taniE: 



Vs s—a 
2r'~2r" 



s—b s — c 
~2r'~2r 




l+ y - 



(s—a) 



12 r 2 



... xf, = A l ■ s 2 + (s-a) 2 + (s-5) 2 + (s-c) 2 
4r 2 \ ^ 12, -2 



, o« + fi* + c» \*_ A A g» + y + c 



4r* 



12 r 5 



4r\ 



(Arts. 101 and 156) 
24 r 2 y 



Er^Afl + ^ + ^Y 

24r 2 ) 



That is : £&e area of the spherical triangle exceeds the area 

a 2 + b 2 + c 2 
o/ £/ie plane triangle by — — — ^- — part of the latter. 

24: Y 

Cor. 2. If we omit terms of the second degree in -, we have 

r 

Er 2 = A. 

Hence, if the sides of a spherical triangle be very small 
compared with the radius of the sphere, its area is approxi- 
mately equal to the area of the plane triangle having sides 
of the same length. 

228. Roy's Rule. — The area of a spherical triangle on the 
Earth's surface being known, to establish a formula for com- 
puting the spherical excess in seconds. 

Let A be the area of the triangle in square feet, and n the 
number of seconds in the spherical excess. Then we have 



BOY'S RULE. 351 



A = Ex60x_60^ Ar 

180° x 60 x 60 v ' 



180 x 60 x 60 206265 



Now, the length of a degree on the Earth's surface is 
found by actual measurement to be 365155 feet. 

,. J* = 365155. ,, r = 18Q x 365155, 

180° it 

Substituting this value of r in (1), and reducing, we get 
log n = log A -9.3267737 (2) 

This formula is called General Roy's rule, as it was used 
by him in the Trigonometric Survey of the British Isles. 
He gave it in the following form : From the logarithm of 
the area of the triangle, taken as a plane triangle, in square, 
feet, subtract the constant logarithm 9.3267737; and the re- 
mainder is the logarithm of the excess above 180°, in seconds, 
nearly. 

Ex. If the observed angles of a spherical triangle are 
42° 2' 32", 67° 55' 39", 70° 1' 48", and the side opposite the 
angle A is 27404.2 feet, required the number of seconds in 
the sum of the errors made in observing the three angles. 
Here the apparent spherical excess is 

A + B + C - 180° = - 1". 
The area of the triangle is calculated from the expression 
a 2 sin B sin C 



2 sin A 



(Art. 101) 



and by Roy's Rule the computed spherical excess is found 
to be .23". 

Xow since the computed spherical excess may be supposed 
to be the real spherical excess, the sum of the observed 
angles ought to have been 180° + .23". 

Hence it appears that the sum of the errors of the obser- 
vations is .23" —(—!") = 1".23, which the observer must 



352 



SPHERICAL TRIGONOMETRY. 



add to the three observed angles, in such proportions as his 
judgment may direct. One way is to increase each of the 
observed angles by one-third of 1".23, and take the angles 
thus corrected for the true angles. 



229. Reduction of an Angle to the Horizon. — Given the 
angles of elevation or depression of two objects, which are at a 
small angular distance from the horizon, and the angle which 
the objects subtend, to find the horizontal angle between them. 

Let a, b be the two objects, the angular distance between 
which is measured by an observer at 
; let OZ be the direction at right 
angles to the observer's horizon. De- 
scribe a sphere round as a centre, 
and let vertical planes through Oa, 
Ob, meet the horizon at OA, OB, re- 
spectively ; then the horizontal angle 
AOB, or AB, is required. 

Let ab = 6, AB = + x, Aa = h, 
B6 = k. Then in the triangle aZb we have 




cos AB = cos aZb ■. 



or 



cos (6 + x) = 



cos ab — cos aZ cos bZ 
sin aZ sin bZ 

cos — sin h sin k 



cos h cos k 



This gives the exact value of AB ; by approximation we 
obtain, where x is essentially small, 

n * cos — lik 

cos 6 — x sin 6 = — • 

l-i(/* 2 + & 2 ) 

.-. x sin = hk — % (h 2 + k 2 ) cos 6, nearly. 



x = 



2 hk - (h 2 + k 2 ) f cos 2 1 - sin 2 ^\ 

2 sin 6 
l[(h + k) 2 tan £ 6 - (h - k) 2 cot £ 0]. 












SMALL VARIATIONS IN PARTS OF TRIANGLES. 353 

EXAMPLES. 

1. Prove that the angles subtended by the sides of a 
spherical triangle at the pole of its circumcircle are respec- 
tively double the corresponding angles of its chordal tri- 
angle. 

2. If A l3 B b Ci ; A 2 , B 2? C 2 ; A 3 , B 3 , C 8 ; be the angles of 
the chordal triangles of the colunars, prove that 

cosA 1 = cos JasinS, cosB 1 = sinJ6sin(S — C), cosC^sin^ csin(S — B), 

cosA 2 = sin Jasin(S — C), cosB 2 = cosJ6sinS, cosC 2 = sin Jcsin(S — A), 

cos A 3 = sin J a sin (S — B) , cos B 3 = sin ^b sin ( S — A ) , cos C 3 = cos \ c sin S. 

3. Prove Legendre's Theorem from either of the formulae 
for sin^-A, cos|-A ? tan -J- A, respectively, in terms of the 
sides. 

4. If C = A + B, prove cos C = — tan \ a tan \ b. 

230. Small Variations in the Parts of a Spherical Tri- 
angle. 

It is sometimes important in Geodesy and Astronomy to 
determine the error introduced into one of the computed 
parts of a triangle from any small error in the given parts. 

If two parts of a spherical triangle remain constant, to deter- 
mine the relation between the small variations of any other two 
parts. 

Suppose C and c to remain constant. 

(1) Required the relation between the small variations 
of a side and the opposite angle (a, A). 

Take the equation 

sin A sine = sin C sin a (1) 

We suppose a and A to receive very small increments da 
and cZA ; then we require the ratio of da and dA when both 
are extremely small. Thus 

sin (A + dA) sin c = sin C sin (a + da), 



354 SPHERICAL TRIGONOMETRY. 

or (sin A cos dA + cos A sin dA) sin c 

= sin C (sin a cos da + cos a sin da) (2) 

Because the arcs dA and da are extremely small, their 
sines are equal to the arcs themselves and their cosines 
equal 1 : therefore (2) may be written 

sin A sin c + cos A sin cd A = sin C sin a + sin C cos ada (3) 

Subtracting (1) from (3), we have 

cos A sin cdA = sin C cos ada. 

da _ cos A sin c _ tan a 
dA sin C cos a tan A 

(2) Required the relation between the small variations 
of the other sides (a, b). We have 

cos c = cos a cos b + sin a sin b cos C (1) 

.\ cosc = cos(a+da)cos(b + db)+sm(a + da)sm(b + db)cosC, 

or = (cos a — sin ada) (cos b — sin bdb) 

+ (sin a + cos ada) (sin b + cos bdb) cos C . (2) 

Subtracting (2) from (1) and neglecting the product 
da db, we have 

= (sin a cos b — cos a sin b cos C) da 

+ (cos a sin b — sin a cos 6 cos C) db, 

_(eot&sina— cosacosC) , (cotasinfr— cos&cosC) „ 
sin a sin b 

__ cot B sin C da cot AsinCdb (\ t IQ^U 

sin a sin 6 

,i da _ cos A 

d& cos B 

(3) Required the relation between the small variations 
of the other angles (A, B). 



SMALL VARIATIONS IN PARTS OF TRIANGLES. 355 

By means of the polar triangle, we may deduce from the 
result just found, that 

dA _ cos a 
clB cos b 

(4) Required the relation between the small variations 
of a side and the adjacent angle (b, A) . We have 

cot c sin b = cot C sin A + cos b cos A . . . (Art. 193) 

Giving to b and A very small increments, and subtracting, 
as before, we get 

cot c cos bdb = cot C cos Ad A — sin b cos Adb — cos b sin A dA. 

(cot c cos b + sin b cos A) db = (cot C cos A — cos b sin A) dA. 

... 22^ db = - 5^ da . . (Arts. 191 and 192) 
sin c sin C 

db _ cos B sin b _ sin b cot B 
dA cos a sin B cos a 

EXAMPLES. 

1. If A and c are constant, prove the following relations 
between the small variations of any two parts of the other 
elements : 

da _ _ tan a t db _ sin a 

dC ~ ~ tan C ' dB~sinC* 

db tan a rZC 

■ — = : — = — cos a. 

dC sin C dB 

2. If B and C remain constant, prove the following : 

db tan b dA . 7 • -, 

— = ; — = sin fr sin C. 

dc tan c da 

dA . A , , da sin a 

- = sin A tan o : 



dc dc sin c cos b 



356 



SPHERICAL TRIGONOMETRY. 



POLYEDRONS. 




231. To find the Inclination of Two Adjacent Faces of a 
Regular Polyedron. 

Let C and D be the centres of the cir- 
cles inscribed in the two adjacent faces 
whose common edge is AB ; bisect AB 
in E, and join CE and DE ; CE and DE 
will be perpendicular to AB. .*. Z CED 
is the inclination of the two adjacent 
faces, which denote by I. 

In the plane CED draw CO and DO at 
right angles to CE and DE, respectively, 
and meeting in 0. Join OA, OE, OB, and from as centre 
describe a sphere, cutting OA, OC, OE at a, c, e, respec- 
tively ; then ace is a spherical triangle. Since AB is per- 
pendicular to CE and DE, it is perpendicular to the plane 
CED ; therefore the plane AOB, in which AB lies, is per- 
pendicular to the plane CED. .-. Z aec is a right angle. 

Let m be the number of sides in each face, and n the 
number of plane angles in each solid angle. Then 

Zace=ZACV = — = ~, 
2m m 

and Z cae = \Z of the planes OAC and OAD. 

.-. Zcae == — ■«= — 
2 n n 



In the right triangle cae we have 
cos cae = cos ce sin ace. 



But 



cos ce = cos COE = cos ( 

\2 2 

7T • I • 7T 

, cos- = sin -sin — 

2 



. I 

: sin — 
2 



sin- 



n 
I 



m 



cos-cosec— • 
n m 



VOLUME OF A PARALLELOPIPED. 357 

Cor. 1. If r be the radius of the inscribed sphere, and a be 
a side of one of the faces, then 

a , 7r , I 

r = -cot — tan — 

2 m 2 

For, r = OC = CEtanCEO=AEcotACEtanCEO 

a , 7T , I 
= - cot — tan — 

2 m 2 
Cor. 2. IfUbe the circumradius of the poly edr on, then 

E = -tan -tan — 
2 w 2 

Eor, r = A cos aoc = E cot eca cot eac = E cot — cot - • 

m n 

.*. E = -tan- tan — 



Cor. 3. Tfte surface of a regular polyedron, E &em# the 

7 - - ma 2 F , 7r 

number of faces, = cot — 

4 m 

a 2 ir 

Eor, the area of one face = — m cot — .*. etc. 

4 m 

Cor. 4. TAe volume of a regular polyedron 

ma 2 r¥ , ?r 
-cot- 



12 m 

Eor, the volume of the pyramid which has one face of 
the polyedron for base and for vertex 

r ma 2 ,ir , 

= cot — - .-.etc. 

3 4 m 

232. Volume of a Parallelopiped. — To find the volume of 
a parallelopiped in terms of its edges and their inclinations 
to one another. 



358 



SPHERICAL TRIGONOMETRY. 



Let the edges be OA = a, 
OB =b, OC = c, and let the 
inclinations be BOC = a, COA 
= 0, AOB = y . Draw CH 
perpendicular to the face 
AOBE. Describe a sphere 
round as centre, meeting 
OA, OB, OC, OE, in a, b, c, e, 
respectively. 

The volume of the parallelo- 
piped is equal to the area of the base OAEB multiplied by 
the altitude CH ; that is, 

volume = ab sin y . CH = abc sin y sin ce 

where ce is the perpendicular arc from c on ab. 

.\ volume = abc sin y sin ac sin bac 

2n 




= abc sin y sin /3 



sin (3 sin y 



(Art. 186) 
(Art. 195) 



= abc Vl — cos 2 a — cos 2 (3 — cos 2 y + 2 cos a cos /? cos y. 
(7or. 1. T/^e surface of a parallelopiped 

= 2 (6c sin a + ca sin/5 + ab sin y). 
CW. 2. The volume of a tetraedron 

=\abc Vl — cos 2 a — cos 2 /?— cos 2 y +2 cos a cos /? cosy. 

Eor, a tetraedron is one-sixth of a parallelopiped which 
has the same altitude and its base double that of the 
tetraedron. 

233. Diagonal of a Parallelopiped. — To find the diagonal 
of a parallelopiped in terms of its edges, and their mutual 
inclinations. 

Let OD (figure of Art. 232) be a parallelopiped, whose 
edges OA = a, OB = b, OC = c, and their inclinations BOC 
= a, COA = /?, AOB = y ; let OD be the diagonal required, 



TABLE OF FORMULA. 359 

and OE the diagonal of the face OAB. Then the triangle 
OED gives 

OD 2 _ OE 2 + ED 2 + 2 OE . ED cos COE 

= a 2 + b 2 + 2 ab cos y + c 2 + 2 c ■ OE cos COE (1) 

Now, it is clear that OE cos COE is the projection of OE 
on the line OC, and therefore it must be equal to the sum 
of the projections of OB and BE (or of OB and OA), on 
the same line.* 

.*. OE cos COE = b cos a + a cos /?, 
which in (1) gives 
OD 2 = a 2 + b 2 + c 2 + 2bc cos« + 2ca cos /} + 2ab cos y . (2) 

234. Table of Formulae in Spherical Trigonometry. — For 

the convenience of the student, many of the preceding 
formulae are summed up in the following table : 

1. cos c = cos a cos b (Art. 185) 

2. sin b = sin B sin c. 

3. sin a = sin A sin c. 

4. cos C = - cos A cos B (Art. 189) 

5. sin B = sin b sin C. 

6. sin A = sin a sin C. 

7. iHi^ = £!EA = iiM ....;.. (Art. 190) 
sin A sin B sin C 

8. cos a = cos b cos c + sin b sin c cos A . . (Art. 191) 

9. cos b = cos c cos a + sin c sin a cos B. 

10. cos c = cos a cos b + sin a sin b cos C. 

11. cos A = —cos B cos C + sin B sin C cos a (Art. 192) 

12. cos B = — cos C cos A + sin C sin A cos b. 

* From the nature of projections (Plane and Solid Geom., Art. 326). 



360 SPHERICAL TRIGONOMETRY. 

13. cos C = — cos A cos B -J- sin A sin B cos c. 

14. cot a sin b = cot A sin C + cos C cos b . . (Art. 193) 

15. cot a sin c = cot A sin B + cos B cos c. 

16. cot b sin a =cot B sin C + cos C cos a. 

17. cot b sin c = cot B sin A + cos A cos c. 

18. cot c sin a = cot C sin B + cos B cos a. 

19. cot c sin b = cot C sin A + cos A cos b. 

20. sin a cos B = cos b sin c — sin b cos c cos A (Art. 194) 

21. sin a cos C = sin b cos c — cos 6 sin c cos A. 

22. sin b cos A = cos a sin c — sin a cos c cos B. 

23. sin b cos C = sin a cos c — cos a sin c cos B. 

24. sin c cos A = cos a sin 6 — sin a cos 6 cos C. 

25. sin c cos B = sin a cos b — cos a sin b cos C. 



26. sin|A = J sm ( s - 5 ) sin ( s - c ) . . . (Art. 195) 
* sin b sin c 



27. cosiA=J sin8sin < 8 - a >. 
\ sin 6 sin c 



28. tan|A=J sin ( 8 -^ sin ( s ~ c > 
\ sins sin (s — a) 



29 sin A - ^aA^ 11 s sin (s — a) sin (s — 6) sin (s — c) 

sin 6 sine 
2n 
sin b sin c 

where n = Vsin s sin (s — a) sin (s — b) sin (s — c). 



30. smig=J- cosScos ( S - A > . . . (Art. 196) 
2 \ sinBsinC v 



31. co S Xg=J^( S - B )J^^) 
2 \ sin B sin C 



TABLE OF FORMULAE. 361 



32. tan i a = J— ^° 8 S T (S " V 
2 \ C os(S-B)cos(S-C) 

33 S i na ^ 2V-cosScos(S-A)cos(S-B)cos(S-C) 

sin B sin C 

34. tani(A + B) = C0S ^ a ~^ cot^C . . (Art. 197) 

35. tan *(A - B) = S1D * (a " 5) coUC. 

36. tan *(« + &) = C0S ^ A "" B ^ tan^c. 

2V J cos^(A + B) 2 

37. tani(a-6) = sin K^ - B) t i 

2V y sin±(A + B) 2 

38. sin^(A + B)cosic = cosi(a-&)cosiC (Art. 198) 

39. sin £(A — B) sin Jc = sin ^(a — 6) cos J.C- 

40. cos J (A + B) cos ^c = cos £(a + 6) sin JC. 

41. cos £(A — B) sin |c = sin £(a + 6) sin ^0. 



42. tanr = / sin - <*) sin (* - &) sin (* - <0 
A/ sins 

. . . (Art. 215) 



sins 
n 



sins 

43. tanE = — [sin(s— a) + sin(s— b) + sin(s— c) — sins] 

2n 

(Art. 217) 

44. K = area of A = — ttt 2 (Art, 219) 

45. siniE = - 5— -- . . . (Art. 220) 

2 cos J a cos -j o cos \ c 

46. tan \ E = Vtan£ stands— a) tan£ (s — 6) tan J(s— c). 



362 SPHERICAL TRIGONOMETRY. 

EXAMPLES. 

1. Find the time of sunrise at a place whose latitude is 
42° 33' N., when the sun's declination is 13° 28' K 

Arts. 5 h 9 m 13 s . 

2. Find the time of sunset at Cincinnati, lat. 39° 6' N., 
when the sun's declination is 15° 56' S. Ans. 5 h 6 m . 

3. Find the time of sunrise at lat. 40° 43' 48" K, in the 
longest day in the year, the sun's greatest declination being 
23° 27' N. Ans. 4 h 32 m 16 s .4. 

4. Find the time of sunrise at Boston, lat. 42°21'N., 
when the sun's declination is 8° 47' S. Ans. 6 h 14 m . 

5. Find the length of the longest day at lat. 42° 16' 48".3 
X., the sun's greatest declination being 23° 27' N. 

Ans. 15 h 5 m 50 8 . 

6. Find the length of the shortest day at New Bruns- 
wick, N. J., lat. 40° 29' 52". 7 N., the sun's greatest declina- 
tion being 23° 27' S. 

7. Find the hour angle and azimuth of Antares, declina- 
tion 26° 6' S., when it sets to an observer at Philadelphia, 
lat. 39° 57' N. Ans. 4 h 23 m 5 8 .7 ; S. 54° 58' 44" W. 

8. Find the hour angle and azimuth of the Nebula of 
Andromeda, declination 40° 35' N., when it rises to an ob- 
server at New Brunswick, N. J., lat. 40° 29' 52".7 N. 

9. Find the azimuth and altitude of Regulus, declination 
16° 13' N., to an observer at New York, lat. 40° 42' N., when 
the star is three hours east of the meridian. 

Ana. Azimuth = S. 71° 12' 30" E. ; Altitude = 44° 10' 33". 

10. Find the azimuth and altitude of Fomalhaut, dec- 
lination 30° 25' S., to an observer in lat. 42° 22' N., when the 
star is 2 h 5 m 36 8 east of the meridian. 

Ans. Azimuth = S. 27° 18' 40" E. ; Altitude = 11° 41' 37". 



EXAMPLES. 363 

11. Find the azimuth and altitude of a star to an observer 
in lat. 39° 57' 1ST., when the hour angle of the star is 
5 h 17 m 40 s east, and the declination is 62° 33' K 

Arts. Azimuth == N. 35° 54' E. ; Altitude = 39° 24'. 

12. Find the hour angle (t) and declination (8) of a star 
to an observer in lat. 40° 36' 23".9 1ST., when the azimuth of 
the star is 80° 23' 4".47, and the altitude is 47° 15' 18".3. 

Ans. t = 46° 40' 4".53 ; 3 = 23° 4' 24".33. 

13. Find the distance between Eegulus and Antares, the 
right ascensions being 10 h ra 29M1 and 16 h 20 m 20 s .35, and 
the polar distances 77° 18' 41".4 and 116° 5' 55".5. 

Ans. 99°55'44".9. 

14. Find the distance between the sun and moon when 
the right ascensions are 12 h 39 m 3 S .22 and 6 h 55 m 32 s .73, and 
the declinations 9° 23' 16".7 S. and 22° 50' 21".9 N. 

Ans. 89°52'55".5. 

15. Find the shortest distance on the earth's surface, in 
miles, from New York, lat. 40° 42' 44" K, long. 74° 0' 24" W., 
to San Francisco, lat. 37° 48' K, long. 122° 23' W. 

Ans. 2562 miles. 

16. Find the shortest distance on the earth's surface 
from San Francisco, lat. 37° 48' K, long. 122° 23' W., to 
Port Jackson, lat. 33° 51' S., long. 151° 19' E. 

Ans. 6444 nautical miles. 

17. Given the right ascension of a star 10 h l m 9 ? .34, and 
its declination 12° 37' 36".8 1ST. ; to find its latitude and longi- 
tude, the obliquity of the ecliptic being 23° 27' 19"-4P 

Ans. Latitude == ; Longitude = 

18. Given the obliquity of the ecliptic <*>, and the sun's 
longitude A ; to find his right ascension a and declination 8. 

Ans. tan a = cos w tan A ; sin 8 = sin w sin A. 



364 SPHERICAL TRIGONOMETRY. 

19. Given the obliquity of the ecliptic 23° 27' 18".5, and 
the sun's longitude 59°40'1".6; to find his right ascension 
(a), and declination (8). 

Am. a = 3 h 49 m 52 8 .62 ; 8 = 20° 5' 33".9 K 

20. Given the sun's declination 16° r 56".4 1ST., and the 
obliquity of the ecliptic 23° 27' 18".2; to find his right 
ascension (a), and longitude (A). 

Ans. a = 9 h 14 m 19 s .2 ; A = 136° V 6".5. 

21. Given the sun's right ascension 14 h 8 m 19 s .06, and the 
obliquity of the ecliptic 23° 27' 17".8 ; to find his longitude 
(A), and declination (8). 

Ans. A = 214° 20' 34".7 ; 8 = 12° 58' 34".4 S. 

22. Given the sun's longitude 280° 23' 52".3, and his 
declination 23° 2' 52".2 S. ; to find his right ascension (a). 

Ans. <* = 18 h 45 m 14 s .7. 

23. In latitude 45° N., prove that the shadow at noon of 
a vertical object is three times as long when the sun's dec- 
lination is 15° S. as when it is 15° N". 

24. Given the azimuth of the sun at setting, and also at 
6 o'clock ; find the sun's declination, and the latitude. 

25. If the sun's declination be 15° N., and length of day 
four hours, prove tan <j> = sin 60° tan 75°. 

26. Given the sun's declination and the latitude ; show 
how to find the time when he is due east. 

27. If the sun rise northeast in latitude <£, prove that cot 
hour angle at sunrise = — sin <fi. 

28. Given the latitudes and longitudes of two places ; 
find the sun's declination when he is on the horizon of both 
at the same instant. 

29. Given the sun's declination 8, his altitude h at 
6 o'clock, and his altitude h' when due east; prove 
sin 2 8 = sin h sin h'. 



EXAMPLES. 365 

30. Given the declination of a star 30° ; find at what 
latitude its azimuth is 45° at the time of rising. 

31. Given the sun's declination S, and the latitude of the 
place <£ ; find his altitude when due east. 

32. Given the declinations of two stars, and the differ- 
ence of their altitudes when they are on the prime vertical ; 
find the latitude of the place. 

33. If the difference between the lengths of the longest 
and shortest day at a given place be six hours, find the 
latitude. 

34. If the radius of the earth be 4000 miles, what is the 
area of a spherical triangle whose spherical excess is 1° ? 

35. If A", B", C" be the chordal angles of the polar 
triangle of ABC, prove 

cos A" = sin % A cos (s — a) , etc. 

36. If the area of a spherical triangle be one-fourth the 
area of the sphere, show that the bisector of a side is the 
supplement of half that side. 

37. If the area of a spherical triangle be one-fourth the 
area of the sphere, show that the arcs joining the middle 
points of its sides are quadrants. 

38. Given the base and area; show that the arc joining 
the middle points of the sides is constant ; and if it is a 
quadrant, then the area of the triangle is 7rr 2 . 

39. Two circles of angular radii, a and /?, intersect 
orthogonally on a sphere of radius r; find in any manner 
the area common to the two. 

40. If E be the spherical excess of a triangle, prove that 
■|E = tan^-a tanifr sinC — ^ (tan J a tanJ6) 2 sin2C + etc. 

41. Show that the sum of the three arcs joining the 
middle points of the sides of the colunars is equal to two 



366 SPHERICAL TRIGONOMETRY. 

right angles, the sides of the original triangle being regarded 
as the bases of the colunars. 

42. Prove that 

cos 2 £a sin 2 S + sin 2 i& sin 2 (S - C) + sin 2 |c sin 2 (S - B) 
+ 2cos^asin|& sin \c sin S sin(S — B) sin(S — C) = 1. 

43. Having given the base and the arc joining the middle 
points of the colunar on the base, the circumcircle is fixed. 

44. Prove sin \ b sin \ c sin (S — A) + cos \ b cos \ c sin S 

= cos^a. 

45. If A + B + C = 2 7T, prove that 

cos 2 ia + cos 2 i& + cos 2 |-c = 1, 
and cos C = — cot \ a cot 1 6. 

46. Solve the equations, 

sin b cos c sin Z + sin c cos b sin Y = sin a, 
sin c cos a sin X + sin a cos c sin Z = sin b, 
sin a cos b sin Y + sin b cos a sin X = sin c, 
for sin X, sin Y, and sin Z. 

47. If b and c are constant, prove the following relations 
between the small variations of any two parts of the other 
elements of the spherical triangle ABC : 

dB tanB da , ^ 

— — = -: = — sin a tan C; 

dC tanC dB 

da • -o dA sin A 

- = sin B sin c ; 



dA. dB sin B cos C 

da • , -o dA sin A 

— = — sin a tan B : — = — • 

dC dC cos B sin C 

48. If A and c remain constant, prove the following : 
da sin a da 



dB tanC' db 



= cos C. 



EXAMPLES. 367 

49. If B and C remain constant, prove the following : 

dA * . da sin a 

— == sin A tan c ; 



db db sin b cos c 

50. If A and a remain constant, prove the following : 
db _ tan b dc __ tan c o 

dB ~~ tan B ' dC ~~ tan C ' 

db cos B d6 sin b 



dc cos C ? dC tan B cos c 

51. Two equal small circles are drawn touching each 
other; show that the angle between their planes is twice 
the complement of their spherical radius. 

52. On a sphere whose radius is r a small circle of spher- 
ical radius 6 is described, and a great circle is described 
having its pole on the small circle ; show that the length 

2 r 

of their common chord is V — cos 2 0. 

sin 6 # 

53. Given the base c of a triangle, and that 

tan \ a tan \ b = tan 2 ^ B, 
B being the bisector of the base, find a — b in terms of c. 

54. If C = A + B, show that 

1 — cos a — cos b + cos c = 0. 

55. If A denote one of the angles of an equilateral tri- 
angle, and A' an angle of its polar triangle, show that 

cos A cos A' = cos A + cos A'. 

56. Show that 

cos a cos B — cos b cos A _ cos C + cos c 
sin a — sin b sin c 

K rr -d ^ ^ a cos & sin b — sin a COS 6 COS C 

57. .rrove cos A = : > 

sin c 

j a . t> 2 sin (a + b) sin 2 1C 
and cos A + cos B = ^ — ' — } - 2 — . 

sine 



368 SPHERICAL TRIGONOMETRY. 



30 ; 

2Y 



58. Prove Legendre's Theorem by means of the relations 

sin A _ sin B _ sin C 

sin a sin b sin c 

59. Two places are situated on the same parallel of lati- 
tude <j> ; find the difference of the distances sailed over by 
two ships passing between them, one keeping to the great 
circle course, the other to the parallel ; the difference of 
longitude of the places being 2 A. 

Arts. 2 r [A cos <£ — sin -1 (co^ <j> sin X) ] . 

60. If the sides of a triangle be each 60°, show that the 
circles described, each having a vertex for pole, and passing 
through the middle points of the sides which meet at it, 
have the sides of the supplemental triangle for common 
tangents. 1 

61. Find the volume and also the inclination of two 
adjacent faces ^1) of a regular tetraedron, (2) of a regular 
octaedron, (3) of a regular dodecaedron, and (4) of a regular 
icosaedron, the edge being one inch. 

Ans. (1) 117.85 cu. in., 70°31'43".4; 

(2) .4714 cu. in., 109° 28' 16"; 

(3) 7.663 cu. in., 116° 33' 54"; 

(4) 2.1817 cu. in., 138° 11' 22".6. 

62. In the tetraedron, prove (1) that the cir<~ jnradius 
is equal to three times its in-radius, and (2) tha , the radius 
of the sphere touching its six edges is a mean proportional 
between the in-radius and circumradius. 

63. Prove that the ratio of the in-radius to the circum- 
radius is the same in the cube and the octaedron, and also 
in the dodecaedron and icosaedron. 



